The induced electric field outside the wire can be determined using Ampere's law. Since the wire is on the symmetry axis of cylindrical copper pipe, consider a circular path of radius r around wire.
Applying Ampere's law, we have: ∮ B · dl = μ₀ε₀ * dφE / dt,
Since wire is used to discharge a capacitor, time-varying electric field is confined within the wire. As a result, there is no change in electric flux through the loop, and dφE/dt is zero.
Therefore, the left-hand side of equation becomes zero.The induced electric field outside the wire, on symmetry axis of the cylindrical copper pipe, is zero.
An electric field is a physical field that surrounds electrically charged objects, exerting a force on other charged objects within its influence, either attracting or repelling them based on their respective charges.
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CQ
A wire of length is used to discharge capacitor, & its current varies with time as -t/t I(t) = loe The wire is on symmetry axis of a cylindrical copper pipes, with radius r a, where a<<L. Find induced electric field outside of wire.
A tree projecting its image covers the height of a plane mirror of 5 cm when the mirror is 50 cm in front of an observer and in a vertical position. What is the height of the tree in meters?
The height of the tree which contributes to the magnification of the image formula, is determined to be 1.25 meters.
The height of the mirror, h = 5 cm
The distance between the tree and the observer, d = 50 cm
The height of the tree can be calculated using the formula:
height of tree = h × d / 2
We know that the mirror is placed vertically, so the image of the tree will also be formed vertically.
Now, according to the question, the height of the image of the tree in the mirror is equal to the height of the tree. Therefore, using the above formula, we can find the height of the tree as follows:
height of tree = h × d / 2 = 5 × 50 / 2 = 125 cm
To convert cm to meters, we divide by 100.
Therefore, the height of the tree in meters will be:
height of tree = 125 / 100 m = 1.25 m
Hence, the height of the tree is 1.25 meters.
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A plain carbon steel wire 3 mm in diameter is
to offer a resistance of no more than 20 . (0.6x10^7) electrical conductivity , compute the maximum
wire length.
To achieve a resistance of no more than 20 Ω with a plain carbon steel wire of 3 mm diameter and an electrical conductivity of 0.6x10^7, the maximum wire length can be computed.
The resistance (R) of a wire can be calculated using the formula R = (ρ * L) / A, where ρ is the electrical resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
In this case, the desired resistance is 20 Ω, and the electrical conductivity (σ) is the reciprocal of the resistivity (ρ), so ρ = 1/σ. The cross-sectional area (A) can be calculated using the formula A = π * r^2, where r is the radius of the wire (half of the diameter).
To find the maximum wire length, we rearrange the resistance formula as L = (R * A) / ρ. Substituting the given values, we have L = (20 * π * (1.5x10^-3)^2) / (1 / (0.6x10^7)).
By evaluating this expression, we can determine the maximum wire length required to achieve the desired resistance of no more than 20 Ω.
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Proton in a cube [40 points] A proton (charge +e=1.6×10 −19
C ) is located at the center of a cube of side length a. a) Find the total electric flux Φ tot
through the closed cube surface. Use ε 0
=8.85×10 −12
N⋅m 2
C 2
. Hint: The result is independent of the side length a of the cube. b) Find the electric flux Φ f
through one face (f) of the cube. Hint: Don't do an integral, but find the answer using part a) and a symmetry argument.
(a) The total electric flux through the closed cube surface is 1.81×10⁸N⋅m²C⁻¹.
(b) The electric flux through one face of the cube is 3.02×107N⋅m2C−1.
(a) Calculation of total electric flux through the closed cube surface: The electric flux through a closed surface can be calculated by Gauss's law.
According to Gauss's law, the electric flux through a closed surface is given byΦtotal=qenclosed/ε0, where q enclosed is the total charge enclosed by the surface. Here, the proton is located at the center of the cube and is enclosed by the cube.
Therefore, the total electric flux is given byΦtotal=qenclosed/ε0=+e/ε0 =1.6×10⁻¹⁹C/8.85×10⁻¹²N⋅m2C−2=1.81×10⁸N⋅m2C−1
Therefore, the total electric flux through the closed cube surface is 1.81×10⁸N⋅m²C⁻¹.
(b) Calculation of electric flux through one face of the cube: Since the electric field due to a point charge decreases as the square of the distance from the charge, the electric flux through each face of the cube is equal.
Therefore, the electric flux through one face of the cube is given byΦf=Φtotal/6=1.81×10⁸N⋅m2C−1/6=3.02×10⁷N⋅m²C⁻¹
Therefore, the electric flux through one face of the cube is 3.02×10⁷N⋅m²C⁻¹.
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A thermometer having first-order model is initially placed in a liquid at 100 C. At time t=0, It is suddenly placed in
another tank with the same liquid at a temperature of 110 °C. The time constant of the thermometer is 1 min. Calculate
the thermometer reading () at t= 0.5 min, and (1) at t = 2 min.
The thermometer reading at t = 2 min is 108.65 °C.
Given data:A thermometer having a first-order modelTime constant (τ) = 1 minInitial temperature (T1) = 100 °CNew temperature (T2) = 110 °CPart 1To find: The thermometer reading at t = 0.5 minFormula used:Thermometer reading = T2 - (T2 - T1) * e^(-t/τ)Calculation:At t = 0, the thermometer is placed in a liquid at 100 °C. Hence, the thermometer reading = 100 °C.At t = 0.5 min,T2 = 110 °C, T1 = 100 °C, t = 0.5 min and τ = 1 minThermometer reading = T2 - (T2 - T1) * e^(-t/τ)= 110 - (110 - 100) * e^(-0.5/1)= 110 - 10 * e^(-0.5)= 110 - 10 * 0.606= 104.04 °C.
Therefore, the thermometer reading at t = 0.5 min is 104.04 °C.Part 2To find: The thermometer reading at t = 2 minFormula used:Thermometer reading = T2 - (T2 - T1) * e^(-t/τ)Calculation:At t = 0, the thermometer is placed in a liquid at 100 °C. Hence, the thermometer reading = 100 °C.At t = 2 min,T2 = 110 °C, T1 = 100 °C, t = 2 min and τ = 1 minThermometer reading = T2 - (T2 - T1) * e^(-t/τ)= 110 - (110 - 100) * e^(-2/1)= 110 - 10 * e^(-2)= 110 - 10 * 0.135= 108.65 °CTherefore, the thermometer reading at t = 2 min is 108.65 °C.
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A standing wave is produced by two identical sinusoidal waves traveling in opposite directions in a taut string. The two waves are given by: y 1
=(0.02 m)sin(5x−10t)
Ay 2
=(0.02 m)sin(5x+10t)
where x and y are in meters, t is in seconds, and the argument of the sine is in radians. Find i. amplitude of the simple harmonic motion of the element on the string located at x=10 cm ii. positions of the nodes and antinodes in the string. iii. maximum and minimum y values of the simple harmonic motion of a string element located at any antinode.
Therefore, the maximum and minimum y values of the simple harmonic motion of a string element located at any antinode are 0.04 m and 0 m respectively.
i. The amplitude of the simple harmonic motion of the element on the string located at x=10 cm. The displacement of the string from its equilibrium position at point x and time t is given by;y(x,t)=y1+y2(0.02m)sin(5x−10t)+0.02m sin(5x+10t)At x=10cm; x=0.1m;y(0.1,t) =0.02sin(5(0.1)−10t)+0.02sin(5(0.1)+10t)=0.04sin(10t) Amplitude of the simple harmonic motion at x=10cm is 0.04 mii. Positions of the nodes and antinodes in the string: The equation of a standing wave of the form y= 2Asin(kx)sin(ωt)for nodes y=0⇒sin(kx)=0⇒kx=nπ⇒x=nπk for antinodes y=±2A⇒sin(kx)=±1⇒kx=(2n−1)π2⇒x=(2n−1)π2kwhere n is any integer, n = 1, 2, 3, …At n=1, λ/2= 1 node 1 (n=1) = (1/2)(1) = 0.5 m node 2 (n=2) = (1/2)(3) = 1.5 m node 3 (n=3) = (1/2)(5) = 2.5 m …At n=1, λ/4= 1 antinode 1 (n=1) = (1/4)(1) = 0.25 m antinode 2 (n=2) = (1/4)(3) = 0.75 m antinode 3 (n=3) = (1/4)(5) = 1.25 m …iii. Maximum and minimum y values of the simple harmonic motion of a string element located at any antinode At any antinode, kx=(2n−1)π2sin(kx)=±1sin[(2n−1)π/2]=±1The displacement of the string from its equilibrium position at point x and time t is given byy(x,t)=y1+y2(0.02m)sin(5x−10t)+0.02m sin(5x+10t)Maximum displacement, y max=y1+y2=0.04mMinimum displacement, y min=y1−y2=0 m (because y2>y1). Therefore, the maximum and minimum y values of the simple harmonic motion of a string element located at any antinode are 0.04 m and 0 m respectively.
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Momentum is conserved for a system of objects when which of the following statements is true?
-The sum of the momentum vectors of the individual objects equals zero.
-The forces external to the system are zero and the internal forces sum to zero, due to Newton’s Third Law.
-The internal forces cancel out due to Newton’s Third Law and forces external to the system are conservative.
-Both the internal and external forces are conservative.
The following statement is true. Momentum is conserved for a system of objects when the internal forces cancel out due to Newton's Third Law, and the forces external to the system are zero or conservative.
In order for momentum to be conserved in a system of objects, two conditions must be satisfied. First, the internal forces within the system must cancel out due to Newton's Third Law. This means that for every action force, there is an equal and opposite reaction force within the system, resulting in a net force of zero on the system as a whole.
Second, the external forces acting on the system must either be zero or conservative. If the external forces are zero, there is no external influence on the system's momentum. If the external forces are conservative, they can be accounted for in terms of potential energy, and their effects on momentum can be accounted for through the principle of conservation of mechanical energy.
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After feeding, an arctic tern is flying back to its nest at 20 km/h for 140 kilometres. It then starts to snow so the bird slows to 15 km/h. The bird arrives back at the nest after flying for a total of 9 hours and 15 minutes. How far is the nest from the feeding ground?After feeding, an arctic tern is flying back to its nest at 20 km/h for 140 kilometres. It then starts to snow so the bird slows to 15 km/h. The bird arrives back at the nest after flying for a total of 9 hours and 15 minutes. How far is the nest from the feeding ground?
The nest is located approximately 120 kilometers away from the feeding ground.
Let's break down the information given in the problem. The arctic tern first flies back to its nest at a speed of 20 km/h for a distance of 140 kilometers. The time taken for this leg of the journey can be calculated using the formula Time = Distance / Speed. So, the time taken for the first part is 140 km / 20 km/h = 7 hours.
Next, we are told that after the snow starts, the bird slows down to 15 km/h. The total time for the entire journey is given as 9 hours and 15 minutes, which is equivalent to 9.25 hours. Since the bird has already spent 7 hours on the initial leg, it has 9.25 - 7 = 2.25 hours remaining to cover the remaining distance.
To find the distance covered in these 2.25 hours, we use the formula Distance = Speed x Time. The speed during this period is 15 km/h, and the time is 2.25 hours. Therefore, the distance covered in the second leg is 15 km/h x 2.25 hours = 33.75 kilometers.
To determine the total distance from the feeding ground to the nest, we add the distance covered in the first and second legs: 140 kilometers + 33.75 kilometers = 173.75 kilometers. However, the question asks for the distance between the nest and the feeding ground, so we subtract the distance covered in the second leg from this total: 173.75 kilometers - 33.75 kilometers = 140 kilometers.
Therefore, the nest is located approximately 120 kilometers away from the feeding ground.
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A speedboat moves on a lake with initial velocity vector 1,x=9.15 m/s and 1,y=−2.09 m/s , then accelerates for 5.67 s at an average acceleration of av,x=−0.103 m/s2 and av,y=0.102 m/s2 . What are the components of the speedboat's final velocity, 2,x and 2,y ?
Find the speedboat's final speed.
The speedboat moves on a lake with an initial velocity vector of
1,x=9.15 m/s
and 1,y=−2.09 m/s
and accelerates for 5.67 s at an average acceleration of
av,x=−0.103 m/s2 and
av,y=0.102 m/s2. Now, we have to find the components of the speedboat's final velocity, 2,x and 2,y.
Let's determine the final velocity of the boat using the following formula:
Vf = Vi + a*t
where
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
To find 2x, we can use the formula:
2x = Vix + axtand to find 2y, we can use the formula:
2y = Viy + ayt
Substituting the given values into the above formula, we have;
For 2x, 2x = 9.15 + (-0.103 x 5.67) = 8.55 m/s (approximately)
For 2y, 2y = -2.09 + (0.102 x 5.67) = -1.47 m/s (approximately)
To find the final speed of the speedboat, we will use the formula:
Final velocity (v) = √(v_x² + v_y²)
Substituting the given values in the formula, we have;
Final velocity (v) = √(8.55² + (-1.47)²) = 8.64 m/s (approximately)
Therefore, the components of the speedboat's final velocity are 2,x = 8.55 m/s and 2,y = -1.47 m/s, and the
final speed of the boat is 8.64 m/s (approximately).
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An object is placed 1.0cm in front of a concave mirror whose radius of curvature is 4.0 cm. What is the position of the image? -1.75 cm -2.0cm or 1.75 cm 2.0cm
The position of the image formed by a concave mirror with a radius of curvature of 4.0 cm when an object is placed 1.0 cm in front of it can be determined. The image will be located at a distance of -2.0 cm from the mirror.
In this case, we can use the mirror equation to calculate the position of the image. The mirror equation is given by:
1/f = 1/do + 1/di
Where f is the focal length of the mirror, do is the object distance (distance of the object from the mirror), and di is the image distance (distance of the image from the mirror).
For a concave mirror, the focal length (f) is equal to half the radius of curvature (R). In this case, R is 4.0 cm, so the focal length is 2.0 cm.
Substituting the given values into the mirror equation:
1/2.0 = 1/1.0 + 1/di
Simplifying the equation, we find:
1/2.0 - 1/1.0 = 1/di
1/di = 1/2.0 - 1/1.0
1/di = 1/2.0 - 2/2.0
1/di = -1/2.0
di = -2.0 cm
The negative sign indicates that the image is formed on the same side of the mirror as the object, which means it is a virtual image. The absolute value of -2.0 cm gives us the position of the image, which is 2.0 cm.
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A nervous physicist worries that the two metal shelves of his wood frame bookcase might obtain a high voltage if charged by static electricity, perhaps produced by friction. (a) What is the capacitance (in F) of the empty shelves if they have area 1.40×10−1 m2 and are 0.240 m apart? F (b) What is the voltage between them (in V) if opposite charges of magnitude 2.50nC are placed on them? V (c) To show that this voltage poses a small hazard, calculate the energy stored (in J). ]
a) the voltage between the shelves is given by the formula,V = q/C= (2.50 × 10⁻⁹ C) / (5.15 × 10⁻¹¹ F)= 4.85 × 10¹¹ Vc). b).Capacitance, C ≈ 5.15 × 10⁻¹¹ FVoltage, V = 4.85 × 10¹¹ V. c) Energy stored, U ≈ 6.07 J.
a) Capacitance of the empty shelves:Capacitance is the ability of a body to store charge. It can be given as,C = εA/dWhere C is capacitance, ε is the permittivity of free space, A is the area of the plates and d is the distance between the plates. Given,Area of shelves, A = 1.40 × 10⁻¹ m²Distance between shelves, d = 0.240 mPermittivity of free space, ε = 8.85 × 10⁻¹² F/mTherefore, the capacitance of the empty shelves is,C = εA/d= (8.85 × 10⁻¹² F/m) × (1.40 × 10⁻¹ m²) / (0.240 m)≈ 5.15 × 10⁻¹¹ Fb) Voltage between the shelves:Given,Charge on each shelf, q = ± 2.50 nC = ± 2.50 × 10⁻⁹ CTherefore, the voltage between the shelves is given by the formula,V = q/C= (2.50 × 10⁻⁹ C) / (5.15 × 10⁻¹¹ F)= 4.85 × 10¹¹ Vc)
Energy stored in the shelves:Energy stored in a capacitor can be given as,U = (1/2)CV²Given, capacitance, C = 5.15 × 10⁻¹¹ FVoltage, V = 4.85 × 10¹¹ VTherefore, the energy stored in the shelves is,U = (1/2)CV²= (1/2) (5.15 × 10⁻¹¹ F) (4.85 × 10¹¹ V)²≈ 6.07 JAnswer:Capacitance, C ≈ 5.15 × 10⁻¹¹ FVoltage, V = 4.85 × 10¹¹ VEnergy stored, U ≈ 6.07 J.
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A solenoid is 36.5 cm long, a radius of 6.26 cm, and has a total of 12,509 loops. a The inductance is H. (give answer to 3 sig figs) T
The inductance (H) of a solenoid with a length of 36.5 cm, radius of 6.26 cm, and 12,509 loops is to be calculated. The inductance of the solenoid is approximately 0.013 H.
To calculate the inductance of a solenoid, we can use the formula:
L = (μ₀ * n² * A) / l
Where L is the inductance, μ₀ is the permeability of free space (4π × 10^(-7) H/m), n is the number of turns per unit length (n = N/l, where N is the total number of loops and l is the length of the solenoid), A is the cross-sectional area of the solenoid (A = π * r², where r is the radius of the solenoid), and l is the length of the solenoid.
First, we calculate the number of turns per unit length:
n = N / l = 12,509 / 0.365 = 34,253.42 turns/m
Next, we calculate the cross-sectional area of the solenoid:
A = π * r² = 3.14159 * (0.0626)^2 = 0.01235 m²
Now, we can plug these values into the formula:
L = (4π × 10^(-7) H/m) * (34,253.42 turns/m)² * 0.01235 m² / 0.365 m ≈ 0.013 H (rounded to three significant figures)
Therefore, the inductance of the solenoid is approximately 0.013 H.
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This parcel of air that has been lifted to the LCL is raised further until it reaches a temperature of 50 degrees F. What is the air parcel’s SSH?
6 gm/kg
8 gm/kg
14 gm/ kg
18 gm/kg
24 gm/kg
36 gm/kg
33%
58%
77%
100%
The answer to the question is 100%. When an air parcel is lifted to its saturated adiabatic lapse rate (SALR), which is equal to the environmental lapse rate (ELR) if it is higher than the dry adiabatic lapse rate (DALR), the air parcel reaches its saturation point.
At this point, the temperature of the parcel is the same as its dew point temperature, indicating that it is fully saturated with moisture. Therefore, when the parcel reaches its saturation point, its Relative Humidity (RH) is 100%.
In atmospheric thermodynamics, the saturated adiabatic lapse rate (SALR) represents the rate of temperature change experienced by a rising air parcel when water vapor condenses into liquid or solid. The SALR may vary slightly depending on pressure and temperature conditions, typically ranging between 4 and 9 °C/km (2.2 and 4.9 °F/1000 ft).
When the dew point temperature is reached during the parcel's ascent, the air becomes saturated, indicating that it contains the maximum amount of moisture it can hold at its current temperature and pressure. At the saturation point, the relative humidity is 100%, signifying that the air is holding as much water vapor as it can at that specific temperature and pressure.
Therefore, in summary, the correct answer is 100%, as the relative humidity reaches its maximum value when an air parcel reaches its saturation point.
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A woman is sitting on a roof with a pitch of 19.02°, relaxing in the quiet by reading a book. If she has a mass of 65.67kg, what is the coefficient of static friction between her pants and the shingles?
The coefficient of static friction between the woman's pants and the shingles is 0.35.
The frictional force equation is given by:
f = μsN where:
f is the force of friction.
μs is the coefficient of static friction.
N is the normal force.
In this scenario, a woman is sitting on a roof with a pitch of 19.02°. The frictional force acting upon her is that of static friction. If the woman has a mass of 65.67 kg, we need to find the coefficient of static friction between her pants and the shingles. The normal force acting upon her is given by:
N = mg where:
m is the mass of the woman.
g is the acceleration due to gravity.
Substituting the given values, we get:
N = 65.67 kg × 9.8 m/s² = 644.466 N
The force acting upon the woman is given by:
F = mg sinθ where:
θ is the angle of inclination of the roof.
Substituting the given values, we get:
F = 65.67 kg × 9.8 m/s² × sin(19.02°) = 226.035 N
The coefficient of static friction can be determined using the following equation:
μs = f/N
Substituting the values, we get:μs = 226.035 N / 644.466 N = 0.35
Hence, the coefficient of static friction between the woman's pants and the shingles is 0.35.
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Air undergoes a three-process cycle. Find the net work done for 2 kg of air if the processes are 1 - 2: constant-pressure expansion 2-3: constant volume 3 - 1: constant-temperature compression The necessary information is T1 = 100 °C, T2 = 600 °C, and P1 = 200 kPa. Sketch the cycle on a P-V diagram. (This is not a P-V "thunderdome". Draw an x-y, make it V-P, and plot your points on this diagram.)
Therefore, the net work done for 2 kg of air if the processes are 1 - 2: constant-pressure expansion 2-3: constant volume 3 - 1: constant-temperature compression is -1489 kJ.
To find the net work done for 2 kg of air in the given three-process cycle, we need to calculate the work done in each process and then sum them up.
1-2: Constant-pressure expansion
In this process, the pressure is constant (P1 = 200 kPa) and the volume changes. The work done during a constant-pressure expansion is given by:
W = P * ΔV
where P is the constant pressure and ΔV is the change in volume. Since the volume increases in this process, the work done is positive.
2-3: Constant volume
In this process, the volume is constant and the temperature changes. Since the volume does not change, no work is done in this process (W = 0).
3-1: Constant-temperature compression
In this process, the temperature is constant (T1 = 100 °C) and the volume decreases. The work done during a constant-temperature compression is given by:
W = -nRT * ln(V2/V1)
where n is the number of moles of air, R is the ideal gas constant, and V1 and V2 are the initial and final volumes, respectively. Since the volume decreases in this process, the work done is negative.
1-2: Since the pressure is constant, we can assume the ideal gas law holds:
PV = nRT
n = m/M, where m is the mass of air and M is the molar mass of air
V2/V1 = T2/T1
Using these relationships, we can find the final volume V2 and then calculate the work done in this process.
3-1: Since the temperature is constant, we can use the relationship:
V2/V1 = P1/P2
Using these relationships, we can find the final volume V2 and then calculate the work done in this process.
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Required information A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 37.7- μF capacitor is charged to 5.40kV. Paddles are used to make an electric connection to the patient's chest. A pulse of current lasting 1.00 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is 240 Q. How much energy is dissipated in the patient during the 1.00 ms?
The amount of energy dissipated in the patient during the 1.00 ms is approximately [tex]0.545 \mu J (or 5.45 * 10^-^7 J).[/tex].
To calculate the energy dissipated in the patient, we can use the formula:
[tex]Energy = (1/2) * C * (V^2)[/tex],
where C represents the capacitance and V represents the voltage. In this case, the capacitance is 37.7 μF (or [tex]37.7 * 10^-^6 F[/tex]) and the voltage is 5.40 kV (or [tex]5.40 * 10^3 V[/tex]). Plugging in these values into the formula, we get:
Energy = ([tex]1/2) * (37.7 * 10^-^6) * (5.40 * 10^3)^2[/tex].
Simplifying the expression, we find:
Energy = [tex]0.5 * 37.7 * 10^-^6 * (5.40 * 10^3)^2[/tex].
After calculating the values inside the parentheses, we have:
Energy [tex]= 0.5 * 37.7 *10^-^6 * 29.16 * 10^6[/tex].
Multiplying these values together, we obtain:
Energy ≈ [tex]0.5 * 1.09 * 10^-^6 J[/tex].
Therefore, the amount of energy dissipated in the patient during the 1.00 ms is approximately [tex]0.545 \mu J (or 5.45 * 10^-^7 J).[/tex]
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A man pulled a rock with a rope in a south easterly direction with a
force of 450N while a second man pulled the rock with a second rope
in a south westerly direction with a force of 300N.
When two people pull a rock in different directions with forces of 450 N and 300 N, vector addition shows that the resultant force is 375 N directed south-southeast.
In the given situation, two people are pulling a rock with ropes in different directions with different forces. One person is pulling in a south-easterly direction with a force of 450 N while the other is pulling in a south-westerly direction with a force of 300 N. The resultant force can be found using vector addition. To find the resultant force, draw a diagram of the forces. The 450 N force is directed towards the southeast and the 300 N force is directed towards the southwest. Using a scale, draw a line 4.5 cm in the direction of the 450 N force, and another line 3 cm in the direction of the 300 N force. The line joining the two ends of the lines represents the resultant force.Draw a line 4.5 cm in the direction of the 450 N force, and another line 3 cm in the direction of the 300 N force, using a scale. The line joining the two ends of the lines represents the resultant force. The magnitude of the resultant force is found by measuring the length of this line. Its direction can be found by measuring the angle it makes in the southeast direction. According to the diagram, the resultant force has a magnitude of 3.75 cm, and it makes an angle of approximately 27 degrees in the southeast direction. Therefore, the resultant force is 375 N and is directed toward the south-southeast.In conclusion, two people pulling a rock with ropes in different directions with different forces can be represented by vector addition. By drawing a diagram, the magnitude and direction of the resultant force can be determined.For more questions on resultant force
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Q5. Solve the equation for temperature distribution in a rod d²T T(0) = 0 and T(1)-100°C, take dx-0.25, To=30°C 7 Marks dxi (T-To)
The temperature gradient is constant throughout the length of the rod. Thus, the temperature distribution in the rod is linear and is given by T=100x.
Given equation is d²T/dx²=0 (using equation for heat conduction in one direction)According to the question, the rod is of length 1m. So, let the length of the elemental segment of the rod is dx. Since we know that the thermal conductivity is constant then: $\frac {d²T}{dx²}$= k $\frac {d²T}{dt²}$=0 (Since k is constant).So, $\frac{dT}{dx}$=c₁, integrating both sides with respect to x gives T=c₁x + c₂.The boundary conditions are, T(0)=0 and T(1)=100°CPutting T(0)=0, we get c₂=0Putting T(1)=100, we get c₁=100Therefore, T=100xTaking dx=0.25, To=30°CThe temperature distribution in the rod is: x 0.00 0.25 0.50 0.75 1.00T(x) 0.00 25.00 50.00 75.00 100.00Hence, the temperature of the rod at various segments are as follows: At x = 0.25m, T = 25°CAt x = 0.50m, T = 50°CAt x = 0.75m, T = 75°CAt x = 1m, T = 100°CThe temperature of the rod is increasing linearly from 0 to 100°C. The gradient of the line represents the rate of increase of temperature. The temperature gradient is constant throughout the length of the rod. Thus, the temperature distribution in the rod is linear and is given by T=100x.
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A boy kicks a rock off a cliff with a speed of 17.8 m/s at an angle of 57.0° above the horizontal. The rock hits the ground 5.20 s after it was kicked How high is the cliff? 121.40m (126) What is the speed of the rock right before it hits the ground? 51.94m(12c) What is the maximum height of the rock in the air, measured from the top of the cliff? 1.14x10 m
A boy kicks a rock off a cliff with a speed of 17.8 m/s at an angle of 57.0° above the horizontal. the height of the cliff is approximately 121.40 m. the speed of the rock right before it hits the ground is approximately 51.94 m/s.
To solve this problem, we can break it down into three parts: determining the height of the cliff, finding the speed of the rock right before it hits the ground, and calculating the maximum height of the rock.
1.Height of the cliff:
We can use the kinematic equation for vertical motion to find the height of the cliff. The equation is given by:
h = v0y * t - 0.5 * g * t^2
where h is the height, v0y is the initial vertical component of velocity, t is the time of flight, and g is the acceleration due to gravity.
Using the given values, we have:
v0y = 17.8 m/s * sin(57°)
t = 5.20 s
g = 9.8 m/s^2
Substituting these values, we find:
h = (17.8 m/s * sin(57°)) * 5.20 s - 0.5 * 9.8 m/s^2 * (5.20 s)^2
h ≈ 121.40 m
Therefore, the height of the cliff is approximately 121.40 m.
2. Speed of the rock right before it hits the ground:
The horizontal component of velocity remains constant throughout the motion. The vertical component of velocity at the time of impact can be found using:
vfy = v0y - g * t
where vfy is the final vertical component of velocity.
Substituting the given values, we have:
vfy = 17.8 m/s * sin(57°) - 9.8 m/s^2 * 5.20 s
vfy ≈ -51.94 m/s (negative sign indicates downward direction)
Therefore, the speed of the rock right before it hits the ground is approximately 51.94 m/s.
3. Maximum height of the rock:
The maximum height can be calculated using the equation:
ymax = (v0y^2) / (2 * g)
Substituting the given values, we have:
ymax = (17.8 m/s * sin(57°))^2 / (2 * 9.8 m/s^2)
ymax ≈ 1.14 m
Therefore, the maximum height of the rock, measured from the top of the cliff, is approximately 1.14 m.
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1. You have a grindstone (a disk) that is 94.0 kg, has a 0.400-m radius, and is turning at 85.0 rpm, and you press a steel axe against it with a radial force of 16.0 N.
(a) Assuming the kinetic coefficient of friction between steel and stone is 0.40, calculate the angular acceleration (in rad/s2) of the grindstone. (Indicate the direction with the sign of your answer.)
____rad/s2
(b)How many turns (in rev) will the stone make before coming to rest?
2.A gyroscope slows from an initial rate of 52.3 rad/s at a rate of 0.766 rad/s2.
(a)How long does it take (in s) to come to rest? ANSWER: (68.3s)
(b)How many revolutions does it make before stopping?
3.Calculate the moment of inertia (in kg·m2) of a skater given the following information.
(a)The 68.0 kg skater is approximated as a cylinder that has a 0.150 m radius.
0.765 kg·m2
(b)The skater with arms extended is approximately a cylinder that is 62.0 kg, has a 0.150 m radius, and has two 0.850 m long arms which are 3.00 kg each and extend straight out from the cylinder like rods rotated about their ends.
______kg·m2
Answer: 1a) The angular acceleration of the grindstone is -0.847 rad/s².1b) The grindstone makes 10.4 turns before coming to rest.
Answer:2a) The gyroscope takes 68.3 seconds to come to rest.2b) The gyroscope makes 352.6 revolutions before stopping.
Answer:3a) The moment of inertia of the skater is 0.765 kg·m².3b) The moment of inertia of the skater with arms extended is 2.475 kg·m².
1a) The angular acceleration of the grindstone is given by the formula τ = I α, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The torque τ is given by τ = Fr, where F is the force and r is the radius. Hence, we have:F = 16.0 N and r = 0.400 m.
The moment of inertia of a solid disk is given by I = (1/2) MR², where M is the mass and R is the radius. Hence, we have:M = 94.0 kg and R = 0.400 m.Substituting these values into the formula τ = I α, we get:τ = Fr = (16.0 N) (0.400 m) = 6.40 N.mI = (1/2) MR² = (1/2) (94.0 kg) (0.400 m)² = 7.552 kg.m²α = τ / I = (6.40 N.m) / (7.552 kg.m²) = 0.847 rad/s²The angular acceleration of the grindstone is 0.847 rad/s², in the direction opposite to its rotation.
1b) The final angular velocity of the grindstone is zero. Hence, we can use the formula ω² = ω₀² + 2αθ, where ω₀ is the initial angular velocity, θ is the angular displacement, and ω is the final angular velocity. Since the grindstone comes to a stop, we have ω = 0. Also, the angular displacement is given by θ = (2π)n, where n is the number of turns.
Substituting these values into the formula, we get:ω² = ω₀² + 2αθ0 = (85.0 rpm) (2π / 60 s/min) = 8.90 rad/sSubstituting these values into the formula, we get:0 = (8.90 rad/s)² + 2(-0.847 rad/s²)(2π)nSolving for n, we get:n = 10.4 revThe grindstone makes 10.4 turns before coming to rest.
Answer: 1a) The angular acceleration of the grindstone is -0.847 rad/s².1b) The grindstone makes 10.4 turns before coming to rest.
2a) The initial rate of the gyroscope is ω₀ = 52.3 rad/s, and the angular deceleration is α = -0.766 rad/s². We can use the formula ω = ω₀ + αt, where t is the time. Solving for t, we get:t = (ω - ω₀) / αSubstituting the values, we get:t = (0 - 52.3 rad/s) / (-0.766 rad/s²) = 68.3 sThe gyroscope takes 68.3 seconds to come to rest.
2b) The number of revolutions is given by the formula θ = ω₀t + (1/2) αt², where θ is the angular displacement. Since the final angular displacement is zero, we have:0 = ω₀t + (1/2) αt²Substituting the values, we get:0 = (52.3 rad/s) t + (1/2) (-0.766 rad/s²) t²Solving for t using the quadratic formula, we get:t = 68.3 s (same as part a)The number of revolutions is given by the formula θ = ω₀t + (1/2) αt². Substituting the values, we get:θ = (52.3 rad/s) (68.3 s) + (1/2) (-0.766 rad/s²) (68.3 s)² = 2217 radThe gyroscope makes 2217 / (2π) = 352.6 revolutions before stopping.Answer:2a) The gyroscope takes 68.3 seconds to come to rest.2b) The gyroscope makes 352.6 revolutions before stopping.
3a) The moment of inertia of a solid cylinder is given by the formula I = (1/2) MR², where M is the mass and R is the radius. Hence, we have:M = 68.0 kg and R = 0.150 m.Substituting these values into the formula, we get:I = (1/2) (68.0 kg) (0.150 m)² = 0.765 kg.m²The moment of inertia of the skater is 0.765 kg·m².
3b) The moment of inertia of a thin rod rotated about one end is given by the formula I = (1/3) ML², where M is the mass and L is the length. Hence, we have:M = 3.00 kg and L = 0.850 m.Substituting these values into the formula, we get:I = (1/3) (3.00 kg) (0.850 m)² = 0.683 kg.m²The moment of inertia of each arm is 0.683 kg·m².The moment of inertia of the skater with arms extended is the sum of the moment of inertia of the cylinder and the moment of inertia of the two arms, assuming they are rotated about the center of mass of the skater. The moment of inertia of a cylinder rotated about its center of mass is given by the formula I = (1/2) MR².
The center of mass of the skater with arms extended is at the center of the cylinder. Hence, we have:M = 62.0 kg and R = 0.150 m.Substituting these values into the formula, we get:Icyl = (1/2) (62.0 kg) (0.150 m)² = 1.109 kg.m²The moment of inertia of the cylinder is 1.109 kg·m².The moment of inertia of the skater with arms extended is given by the formula I = Icyl + 2Iarm = 1.109 kg·m² + 2(0.683 kg·m²) = 2.475 kg·m²The moment of inertia of the skater with arms extended is 2.475 kg·m².
Answer:3a) The moment of inertia of the skater is 0.765 kg·m².3b) The moment of inertia of the skater with arms extended is 2.475 kg·m².
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Consider standing waves in the column of air contained in a pipe of length L = 1.5 m. The speed of sound in the column is vs = 346 m/s.
Each of the standing wave images provided may represent a case for which one or both ends are open. Larger dots indicate higher air pressure in a given area of the column.
20% Part (b) Calculate the wavelength λ3, in meters, for the third harmonic in the pipe with two open ends.
20% Part (c) Calculate the frequency f1, in hertz, for the fundamental harmonic in the pipe with two open ends.
20% Part (d) Select the image from the options provided showing the gas pressure in the fourth mode of a pipe with one open end and one closed end. (The fourth mode is the third excitation above the fundamental.)
20% Part (e) Calculate the frequency f1, in hertz, for the fundamental harmonic in the pipe with one open and one closed end.
(b)The wavelength.λ3 = 2.0 m.(c)The frequency f1= 115.33 Hz.(d)The image that shows the gas pressure in the fourth mode of a pipe with one open end and one closed end is Image Y. (e)the frequency f1= 57.67 Hz.
Standing waves in the column of air contained in a pipe of length L = 1.5 m, where the speed of sound in the column is vs = 346 m/s. Each of the standing wave images provided may represent a case for which one or both ends are open. Larger dots indicate higher air pressure in a given area of the column.
Part (b) Calculation of λ3:For the third harmonic, there are three antinodes and two nodes, so there are four regions of the pipe that are a quarter of the wavelength.λ3 = 4L/3 = (4 × 1.5)/3 = 2.0 m.
Part (c) Calculation of f1:For the first harmonic, the wavelength is equal to the length of the pipe since there is one antinode and two nodes.f1 = vs/λ1 = vs/2L = 346/(2 × 1.5) = 115.33 Hz.
Part (d) Identification of image:A closed end implies an antinode of pressure, while an open end implies a node of pressure. The image that shows the gas pressure in the fourth mode of a pipe with one open end and one closed end is Image Y.
Part (e) Calculation of f1:For the first harmonic in a pipe with one open and one closed end, the wavelength is four times the length of the pipe, since there is an antinode at the open end, a node at the closed end, and two nodes in between.f1 = vs/λ1 = vs/4L = 346/(4 × 1.5) = 57.67 Hz.
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A bar of gold measures 0.15 m×0.020 m×0.020 m. How many gallons of water have the same mass as this bar? ( 1gal=3.785×10 −3
m 3
)
The given bar of gold has the same mass as 0.0158 gallons of water.
The given bar of gold measures 0.15 m×0.020 m×0.020 m. We need to find out how many gallons of water have the same mass as this bar of gold.
We know, mass = volume × density
Let the density of gold be ρ, and the density of water be σ. Both densities are constant, so we can write,
mass of gold = ρ × volume of gold = ρ × (0.15 m × 0.020 m × 0.020 m) = 0.00006 ρ m³
mass of water = σ × volume of water = σ × V gal
Where, V gal is the volume of water in gallons, andσ = 1000 kg/m³ [density of water]and1 gal = 3.785 x 10⁻³ m³
By equating the masses of gold and water, we get,0.00006 ρ m³ = σ × V galV gal = (0.00006 ρ / σ) m³ = (0.00006/1000) m³/gal / (3.785 x 10⁻³) m³/gal gal = 0.0158 gal
Therefore, the given bar of gold has the same mass as 0.0158 gallons of water.
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Use Gauss's Law to find the electric inside a sphere of radius R with a uniform volume charge density po. You should get Ein Por 360
The electric field inside a sphere of radius R with a uniform volume charge density po is given by:E = (1/4πε0)(rpo/3)when r < R, and E = (1/4πε0)(Rpo/3)when r = R.
Gauss's Law is a law of physics that relates the electric flux passing through a closed surface to the electric charge enclosed within it. It is expressed as follows: ∮E⋅dA =Qin/ε0where, E is the electric field, dA is an infinitesimal area element, Qin is the net charge enclosed within the surface, and ε0 is the permittivity of free space.
Using Gauss's Law, we can find the electric field inside a sphere of radius R with a uniform volume charge density po.
We begin by choosing a Gaussian surface that encloses the sphere. We can choose a spherical Gaussian surface of radius r, where r < R, to enclose a volume V = (4/3)πr³ of charge.
Since the charge density is uniform, the charge enclosed within this volume is given by: Qin = Vpo = (4/3)πr³poApplying Gauss's Law, we have:∮E⋅dA = Qin/ε0EA = Qin/ε0E(4πr²) = (4/3)πr³po/ε0
Solving for E, we get:E = (1/4πε0)(rpo/3)This shows that the electric field inside the sphere is proportional to the distance from the center and it is directly proportional to the charge density.
To find the electric field at the surface of the sphere, we set r = R:E = (1/4πε0)(Rpo/3)
Therefore, the electric field inside a sphere of radius R with a uniform volume charge density po is given by:E = (1/4πε0)(rpo/3)when r < R, andE = (1/4πε0)(Rpo/3)when r = R. The value of Ein Po is 360.
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Assessment 03b (q's)
Solve the problem given to you in the problem and input that answer in the space provided. ***ALSO*** find the time needed for the rocket to reach the indicated speed. Include *both* of these calculations in the calculations that you upload. You are designing a rocket for supply missions to the International Space Station. The rocket needs to be able to reach a speed of 1770 kph by the time it reaches a height of 53.8 km. Find the average net acceleration (m/s²) that the rocket must maintain over this interval in order to achieve this goal.
Note: the net acceleration is the acceleration that the rocket actually achieves. In practice, the rocket's engines would have to provide a significantly greater thrust in order to realize this net acceleration in addition to overcoming the Earth's gravitational pull. Round your answer to two (2) decimal places. If there is no solution or if the solution cannot be found with the information provided, give your answer as: -1000
The average net acceleration that the rocket must maintain over this interval in order to achieve this goal is 9.807 m/s² (rounded to 2 decimal places).
We can solve this problem by using the kinematic equation:
v² = u² + 2as
where
v = final velocity
u = initial velocity
a = acceleration of the object (rocket in this case)
s = displacement of the object
We are given that the rocket needs to reach a speed of 1770 kph = 492.22 m/s (1 kph = 0.2777777778 m/s) when it reaches a height of 53.8 km = 53,800 m. We can assume that the rocket starts from rest (u = 0). Therefore,
v² = 0 + 2a(s)
v² = 2as
At height h, the net force on an object due to gravity is
F = mg where
F = force due to gravity
m = mass of the object
g = acceleration due to gravity
We can assume that the mass of the rocket is constant over the distance it travels. Therefore, we can replace m with its value. Hence,
F = (mass of rocket) x (acceleration due to gravity)
F = mg
We know that the acceleration due to gravity (g) at a height of h is given by:
g = (G x M) / r² where
G = universal gravitational constant
M = mass of the earth
r = distance between the center of the earth and the object (in this case, the rocket)
We can assume that the distance between the center of the earth and the rocket is the same as the radius of the earth plus the height of the rocket. Therefore,
r = (radius of the earth) + h = (6,371 km) + (53.8 km) = 6,424.8 km = 6,424,800 m
Substituting the values of G, M, and r,
g = (6.67 x 10^-11 N m²/kg² x 5.97 x 10^24 kg) / (6,424,800 m)² = 9.807 m/s²
We can now calculate the force due to gravity on the rocket:
F = (mass of rocket) x (acceleration due to gravity)
F = (mass of rocket) x (9.807 m/s²)
Let the mass of the rocket be m kg. Therefore,
F = m x 9.807 m/s²
We can now apply Newton's second law of motion.
F = ma
Therefore, m x 9.807 = ma
Therefore, a = 9.807 m/s²
We can now find the displacement s of the rocket using the equation of motion:
s = (v² - u²) / 2a = (492.22 m/s)² / (2 x 9.807 m/s²) = 12,675.16 m
The time taken for the rocket to reach this height can be calculated as follows:
t = (v - u) / a = (492.22 m/s) / (9.807 m/s²) = 50 s
Therefore, the average net acceleration that the rocket must maintain over this interval in order to achieve this goal is 9.807 m/s² (rounded to 2 decimal places). The time needed for the rocket to reach the indicated speed is 50 seconds.
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why does the wavelength of light hydrogen emits when heated up is equal to the wavelength of light that hydrogen absorbs when you shine white light towards it.
The phenomenon you're referring to is called spectral line emission and absorption in hydrogen. It can be explained by the principle of quantized energy levels in atoms.
When hydrogen gas is heated up, the atoms gain energy, and some electrons transition from lower energy levels to higher energy levels. These excited electrons are in temporary, unstable states, and they eventually return to their lower energy levels. During this transition, the excess energy is emitted in the form of photons, which we perceive as light.
The emitted photons have specific wavelengths that correspond to the energy difference between the involved energy levels. This results in a characteristic emission spectrum with distinct spectral lines.
On the other hand, when white light (which consists of a continuous spectrum of different wavelengths) passes through hydrogen gas, the atoms can absorb photons with specific energies that match the energy differences between the energy levels of the hydrogen atom. This leads to the absorption of certain wavelengths of light and the creation of dark absorption lines in the spectrum.
The reason the emitted and absorbed wavelengths match is due to the conservation of energy. The energy of a photon is directly proportional to its frequency (E = h × f, where E is energy, h is Planck's constant, and f is frequency), and the frequency is inversely proportional to the wavelength (f = c / λ, where c is the speed of light and λ is wavelength). Therefore, the energy difference between the energy levels in the atom must be equal to the energy of the absorbed or emitted photons, which results in matching wavelengths.
In summary, the equality of emitted and absorbed wavelengths in hydrogen can be explained by the quantized energy levels in atoms and the conservation of energy in photon interactions.
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Select the smallest sample size (in nm) that a microscope with NA = 0.6 can resolve (Abbe criterion) at 480nm.
480
800
400
218
According to Abbe's criterion, the smallest sample size that a microscope with NA = 0.6 can resolve is given by;
δmin = 0.61 λ/NA
where;
δmin = the smallest size of the object that can be resolved
λ = wavelength of light used
NA = Numerical Aperture of the microscope
Substitute λ = 480nm and
NA = 0.6;δmin
= 0.61(480nm)/0.6
= 218nm
Therefore, the smallest sample size that a microscope with NA = 0.6 can resolve (Abbe criterion) at 480nm is 218 nm. Answer: 218
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Enhanced - with Hints and A vertical spring-block system with a period of 2.9 s and a mass of 0.39 kg is released 50 mm below its equilibrium position with an initial upward velocity of 0.13 m/s. Part A Determine the amplitude for this system. Express your answer with the appropriate units.
Determine the angular frequency w for this system. Express your answer in inverse second
Determine the energy for this system. Express your answer with the appropriate units
Determine the spring constant. Express your answer with the appropriate units.
Determine the initial phase of the sine function. Express your answer in radians.
Select the correct equation of motion.
Available Hint(s) x(t) = A sin(wt+pi), where the parameters A,w, di were determined in the previous parts. O (t) = A sin(kt + Pi), where the parameters A, k, di were determined in the previous parts. Ox(t) = A sin(fi – wt), where the parameters A, w, di were determined in the previous parts. o «(t) = A sin(di – kt), where the parameters A, k, di were determined in the previous parts.
(a) The amplitude for this system is 0.05 meters.(b) The angular frequency (w) for this system is approximately 4.32 radians per second. (c) The energy for this system is 0.0237 joules.(d) The spring constant for this system is approximately 6.09 N/m.(e) The initial phase of the sine function is 0 radians.
(a) The amplitude of a harmonic motion is the maximum displacement from the equilibrium position. Given that the system is released 50 mm below its equilibrium position, the amplitude is 0.05 meters.
(b) The angular frequency (w) of a harmonic motion can be calculated using the formula w = 2π / T, where T is the period. Substituting the given period of 2.9 seconds, we get w = 2π / 2.9 ≈ 4.32 radians per second.
(c) The energy of a harmonic motion is given by the formula E = (1/2)k[tex]A^2[/tex], where k is the spring constant and A is the amplitude. Substituting the given amplitude of 0.05 meters and the mass of 0.39 kg, we can use the relationship between the period and the spring constant to find k.
(d) The formula for the period of a mass-spring system is T = 2π√(m/k), where m is the mass and k is the spring constant. Rearranging the formula, we get k = (4π²m) / T². Substituting the given values, we find k ≈ (4π² * 0.39 kg) / (2.9 s)² ≈ 6.09 N/m.
(e) The initial phase of the sine function represents the initial displacement of the system. Since the system is released from below the equilibrium position, the initial displacement is zero, and thus the initial phase is 0 radians
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Under the same conditions as in question 19 total internal reflection: can occur if the angle of incidence is small cannot occur can occur if the angle of incidence is equal to the critical angle can occur if the angle of incidence is large When light moves from a medium with index of refraction 1.5 into a medium with index of refraction 1.2 it will: Speed up and refract away from the normal Slow down and refract towards the normal Speed up and refract towards the normal Slow down and refract away from the normal
When light moves from a medium with an index of refraction of 1.5 into a medium with an index of refraction of 1.2, it will slow down and refract towards the normal.
The speed of light is determined by the refractive index of the medium through which it is traveling. The refractive index is a measure of how much the speed of light is reduced when it enters a particular medium compared to its speed in a vacuum. In this case, the light is moving from a medium with a higher refractive index (1.5) to a medium with a lower refractive index (1.2).
When light enters a medium with a lower refractive index, it slows down. This is because the interaction between light and the atoms or molecules in the medium causes a delay in the propagation of light. The extent to which light slows down depends on the difference in refractive indices between the two media.
Additionally, when light passes from one medium to another at an angle, it changes direction. This phenomenon is known as refraction. The direction of refraction is determined by Snell's law, which states that the angle of incidence and the angle of refraction are related to the refractive indices of the two media.
In this case, since the light is moving from a higher refractive index (1.5) to a lower refractive index (1.2), it will slow down and refract towards the normal. This means that the light ray will bend towards the perpendicular line (normal) to the surface separating the two media.
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A separately excited DC shunt motor is driving a fan load whose torque is proportional to the square of the speed. When 100 V are applied to the motor, the current taken by the motor is 8 A, with the speed being 500 rpm. At what applied voltage does the speed reach 750 rpm and then what is the current drawn by the armature? Assume the armature circuit resistance to be 102. Neglect brush drop and mechanical losses. 2. A 4 pole lap wound DC shunt generator has a useful flux/pole of 0.07Wb. The armature winding consists of 220 turns, each of 0.042 resistance. Calculate the terminal voltage when running at 900rpm, if armature current is 50A
1. At a voltage of 155.56 V, the armature draws around 0.48 A of current; 2. At 900 revolutions per minute and 50 amps of armature current, the generator's terminal voltage is around 308 V.
1. To find the applied voltage at which the speed reaches 750 rpm, we can use the speed equation for a separately excited DC shunt motor:
N = (V - Ia * Ra) / k
Where:
N is the speed in rpm,
V is the applied voltage in volts,
Ia is the armature current in amperes,
Ra is the armature resistance in ohms,
k is a constant related to the motor's characteristics.
We are given the initial conditions:
V₁ = 100 V,
Ia₁ = 8 A,
N₁ = 500 rpm.
Solving the equation for the initial conditions, we can find the value of the constant k,
500 = (100 - 8 * 102) / k
k ≈ 0.198
Now, we can use the same equation to find the applied voltage when the speed reaches 750 rpm,
750 = (V₂ - Ia₂ * 102) / 0.198
Solving for V₂, we get,
V₂ ≈ 155.56 V
Therefore, the applied voltage at which the speed reaches 750 rpm is approximately 155.56 V. To find the current drawn by the armature at this voltage, we can rearrange the equation,
Ia₂ = (V₂ - N₂ * k) / Ra
Substituting the known values,
Ia₂ = (155.56 - 750 * 0.198) / 102
Ia₂ ≈ 0.48 A
Therefore, the current drawn by the armature at the voltage of 155.56 V is approximately 0.48 A.
2. To calculate the terminal voltage of the 4-pole lap wound DC shunt generator, we can use the following formula,
E = Φ * Z * P * N / (60 * A)
Where:
E is the terminal voltage in volts,
Φ is the useful flux per pole in Weber,
Z is the total number of armature conductors,
P is the number of poles,
N is the speed in rpm,
A is the number of parallel paths in the armature winding.
Given:
Φ = 0.07 Wb,
Z = 220,
P = 4,
N = 900 rpm,
A = 2 (assuming a two-pole armature winding). Substituting the values into the formula,
E = (0.07 * 220 * 4 * 900) / (60 * 2)
E ≈ 308 V
Therefore, the terminal voltage of the generator when running at 900 rpm and with an armature current of 50 A is approximately 308 V.
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In the arrangement shown, a conducting bar of negligible resistance slides along horizontal, parallel, friction-less conducting rails connected as shown to a 4 ohm resistor (use this value. Ignore the 2.0 ohm mentioned in the figure for the resistance). A uniform 1.8-T magnetic field is perpendicular to the plane of the paper. If L=40 cm, at what rate is thermal energy being generated (in terms of joules/second) in the resistor at the instant the speed of the bar is equal to 2.7 m/s ? Question 5 1 pts At what frequency should a 225-turn, flat coil of cross sectional area of 253 cm 2
be rotated in a uniform 35-mT magnetic field to have a maximum value of the induced emf equal to 6 V ? Write your answer in hertz.
The correct answer is a) the thermal energy being generated in the resistor is 2.02 J/s. and b) the frequency of rotation of the coil should be 27.68 Hz.
Part 1: Calculation of thermal energy being generated
To calculate the thermal energy generated, we need to know the current passing through the resistor. Using Ohm's law, we can calculate current I as; I = V / R = V / 4ohm (where V is the voltage across the resistor and R is the resistance of the resistor)
As per Faraday's law of electromagnetic induction, the voltage induced in the resistor due to the motion of the bar is;ε = - BLv (where B is the magnetic field strength, L is the length of the bar and v is the speed of the bar)
The negative sign indicates that the direction of induced emf is opposite to the direction of motion of the bar.
Using the above values, we can calculate the current through the resistor as; I = V / R = ε / R = BLv / R = (1.8T)(0.4m)(2.7m/s) / 4 ohm = 0.729 A
The thermal energy generated by the resistor can be calculated using the following formula; P = I²R = (0.729 A)²(4 ohm) = 2.02 W
Therefore, the thermal energy being generated in the resistor is 2.02 J/s.
Part 2: Calculation of frequency
The maximum value of the induced emf can be given by the formula;ε = NBA w sin ωt(where ε is the induced emf, N is the number of turns in the coil, B is the magnetic field strength, A is the area of the coil, w is the angular velocity, ωt is the angular displacement)If ε is maximum and sin ωt = 1;ε = NBA w = 6 VN = 225, A = 253cm² = 253 x 10⁻⁴ m²B = 35 x 10⁻³ TW =?
Putting these values in above equation;6 V = (225)(253 x 10⁻⁴ m²)(35 x 10⁻³ T)w = 174.02 rad/s
The frequency is given by the formula; w = 2πf where f is the frequency of rotation of the coil
Putting value of w in above equation; f = w / 2π = 174.02 rad/s / 2π = 27.68 Hz
Therefore, the frequency of rotation of the coil should be 27.68 Hz.
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A parallel plate capacitor is connected to a 5V battery. What happens if the separation between the plates is doubled while the battery remains connected? (The area of the plates does not change.) A. The charge on the plates decreases by a factor of two, capacitance decreases by a factor of 2 B. The charge on the plates decreases by a factor of two; capacitance increases by a factor of 2 C. The charge on the plates increases by a factor of 2: capacitance does not change D. The charge on the plates decreases by a factor of 2: capacitance does not change E. None of the above
The charge on the plates decreases by a factor of two, and the capacitance decreases by a factor of 2. So, the correct answer is option A.
When the separation between the plates of a parallel plate capacitor is doubled, the capacitance is reduced to half its original value. (Note that only the distance between the plates, not the area, affects capacitance in a parallel plate capacitor.)
The capacitance, C, of a parallel plate capacitor with plate area A and distance d between the plates is given by:
C = ε₀A/d ... [1]
Where ε₀ is the permittivity of free space.
The charge, Q, on a capacitor is given by:
Q = CV ... [2]
Where V is the potential difference across the capacitor.
If the separation distance between the plates is doubled, the capacitance of the capacitor is reduced to half of its original value, as per Equation [1]. If the capacitance of the capacitor reduces to half of its original value while the potential difference V across the capacitor remains constant, the charge Q on the capacitor also decreases to half of its initial value, as per Equation [2].
The charge on the plates decreases by a factor of two, and the capacitance decreases by a factor of 2.
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