A closed-loop system has sustained oscillations (i.e., constant amplitude) with a period of 66 seconds when the gains are proportional and integral.
A closed-loop system is a type of control system in which the output is fed back into the system to regulate the input. In this way, the output controls the input. A closed-loop system is a type of feedback control system that includes a controller, a plant, and a sensor. The closed-loop system's sustained oscillations occur due to the interaction between the plant, sensor, and controller. The oscillations are a result of the feedback loop's positive feedback, which amplifies the input and causes it to oscillate.
In a closed-loop system, the gains of the controller play a critical role in determining the oscillations' characteristics. The period of oscillation in a closed-loop system is the time taken for one complete cycle of oscillation. The period of oscillation is measured in seconds. The question states that the closed-loop system has a period of 66 seconds, which implies that it takes 66 seconds to complete one cycle of oscillation. Proportional and integral gains are the two types of gains utilized in a closed-loop system's controller. Proportional gain is the direct relationship between the output and the input. Integral gain, on the other hand, is the error signal's integral over time.
By adding these two gains, the controller may regulate the plant to create the desired output. The fact that the closed-loop system has sustained oscillations with a period of 66 seconds when the gains are proportional and integral implies that the system's feedback loop is stable. In other words, the feedback loop is generating constant oscillations with a fixed period, which is a sign of a stable system.
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brody the dog is sitting at your feet like a good boy. you throw a dog toy away from you at a speed of 34 m/s (good arm). brody runs after the ball at a constant speed of 7 m/s. the toy is designed to make a constant high frequency tone of 950 hz as it flies through the air. what frequency does brody hear initially as he chases the toy?
Brody initially hears the frequency of the toy as 1071 Hz (approximately).
To solve this problem, we need to use the Doppler effect formula, which relates the frequency of a wave to the relative motion between the source and the observer.
The frequency heard by Brody initially:
[tex]f' = f * (v + vo) / (v + vs)[/tex]
where f is the frequency of the toy (950 Hz), v is the speed of sound (assumed to be 343 m/s at room temperature), vo is the velocity of the observer (Brody, running at 7 m/s), and vs is the velocity of the source (the toy, moving away from Brody at 34 m/s).
[tex]f' = 950 Hz * (343 m/s + 7 m/s) / (343 m/s - 34 m/s)\\\f' = 950 Hz * 350 / 309\\\\\f' = 1071 Hz (approximately)[/tex]
Therefore, Brody initially hears the frequency of the toy as 1071 Hz (approximately).
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a musician is tuning her cello (a string instrument) to the key of c, so that the string vibrates at a frequency of 65.4 hz when played. the string is fixed on both ends, is 0.600 m long, and weighs 0.141 n. if she wants to raise the frequency to sound a d (73.4 hz) rather than c, what percentage increase in tension is needed?
The musician needs to increase the tension in the string by approximately 29.9% to raise the frequency from C to D.
To raise the frequency of a string, a musician can change the tension on the string.
The relationship between the frequency of a string and its tension can be described by the following equation:
f = (1/2L) x [tex]\sqrt{(T/\mu)}[/tex]
where f is the frequency,
L is the length of the string,
T is the tension in the string,
and μ is the linear mass density of the string (mass per unit length).
In this scenario, the musician wants to raise the frequency of the string from 65.4 Hz to 73.4 Hz by changing the tension on the string.
The length of the string is fixed at 0.600 m, and the mass of the string is given as 0.141 N.
We can start by using the given values to solve for the initial tension in the string when it is tuned to C.
Rearranging the equation above and plugging in the given values, we get:
T = [tex]\mu \times (2Lf)^2[/tex]
where μ = m/L is the linear mass density of the string,
and f = 65.4 Hz. Plugging in the values, we get:
μ = m/L = 0.141 N / 0.600
m = 0.235 kg/m
T = (0.235 kg/m) [tex](2 \times 0.600 m \times 65.4 Hz)^2[/tex]
= 200.3 N
Now, we want to find the tension in the string that will result in a frequency of 73.4 Hz when the string is played.
Let's call this new tension T'.
Using the same equation as before, we can solve for T':
T' = μ x [tex](2Lf')^2[/tex]
where f' = 73.4 Hz.
We want to find the percentage increase in tension needed to achieve this new frequency, so we can write:
% increase in tension = (T' - T) / T x 100%
Plugging in the values and solving for T', we get:
T' = [tex]\mu \times (2Lf')^2[/tex]
= (0.235 kg/m) x [tex](2 \times 0.600 m \times 73.4 Hz)^2[/tex]
= 260.2 N
So the percentage increase in tension needed is:
% increase in tension = (T' - T) / T x 100%
% increase in tension = (260.2 N - 200.3 N) / 200.3 N x 100% ≈ 29.9%
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a ski starts from rest and slides down a 30o incline 150m long. a) if the coefficient of friction is 0.15, what is the ski's speed at the base of the incline? b) if the snow is level at the foot of the incline and has the same coefficient of friction, how far will the ski travel along the level? use energy methods.
a) Using energy methods and assuming negligible air resistance, the speed of the ski at the base of the incline is approximately 34.1 m/s. b) The ski will travel approximately 110.6 m along the level snow.
a) When the ski descends the gradient, its original potential energy is transformed into kinetic energy. Friction causes part of this energy to be wasted, which reduces the ski's speed at the bottom of the hill. The ski's ultimate kinetic energy may be calculated by subtracting the work done by friction from its starting potential energy using the laws of energy conservation. b) The ski only has kinetic energy left as it reaches the bottom of the gradient since all of its potential energy has been expelled. The ski's kinetic energy is transformed into thermal energy as it travels down the flat snow by the frictional force pressing on it, slowing it down until it ultimately comes to a stop. making use of energy saving once more.
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Photoelectric effect
7.
A metal surface having a work function of 3.0 eV is illuminated with radiation of wavelength
350nm. Calculate:
a) The threshold frequency (fo) and wavelength (Ao)
b) The maximum kinetic energy of the emitted photoelectrons
a) Calculate the work function (in eV) for a magnesium surface if the minimum frequency of
electromagnetic radiation which causes photoemission from the metal surface is
8.9 x 10¹4 Hz. in Joules
b) If the same surface were illuminated with radiation of wavelength 250 nm, calculate:
i. The maximum kinetic energy,
ii. The maximum velocity, of the emitted photoelectrons
9. When electromagnetic radiation of frequency 1.5 x 1014 Hz is incident on a metal surface, the
maximum kinetic energy of the emitted photoelectrons is found to be 3.8 x 10-20 J. Calculate the
work function of the metal.
10. Photons of electromagnetic radiation having energies of 1.0 eV, 2.0 eV and 4.0 eV are incident on a
metal surface having a work function of 1.7 eV.
a) Which of these photons will cause photoemission from the metal surface?
b) Calculate the maximum kinetic energies (in eV and J) of the liberated electrons in each of
those cases where photoemission occurs.
11. A vacuum photocell connected to a microammeter is illuminated with light of varying wavelength.
a) Explain why:
i. A photoelectric current is registered on the microammeter when light of a certain
wavelength is incident on the photocell.
ii. The current is found to increase with the light intensity is increased.
b) When the incident light wavelength is increased, the photoelectric current falls to zero. decre-
ased.
Explain why:
i. The current falls to zero.
ii. The current would still be zero if the light wavelength is kept the same and the
intensity is increased.
Explanation:
7a) The work function (ϕ) is the minimum energy required to remove an electron from the metal surface. It is related to the threshold frequency (fo) by the equation:
ϕ = hfo
where h is Planck's constant (6.626 x 10^-34 J s).
The threshold wavelength (Ao) can be calculated from the threshold frequency using the equation:
c = λf
where c is the speed of light (3.00 x 10^8 m/s).
Given that the work function of the metal surface is 3.0 eV, we have:
ϕ = 3.0 eV = (3.0 x 1.6 x 10^-19) J fo = ϕ/h = (3.0 x 1.6 x 10^-19) J / (6.626 x 10^-34 J s) ≈ 4.53 x 10^14 Hz Ao = c/fo = (3.00 x 10^8 m/s) / (4.53 x 10^14 Hz) ≈ 661 nm
Therefore, the threshold frequency is 4.53 x 10^14 Hz and the threshold wavelength is approximately 661 nm.
7b) The maximum kinetic energy of the emitted photoelectrons can be calculated using the equation:
KEmax = hf - ϕ
where h is Planck's constant, f is the frequency of the incident radiation, and ϕ is the work function of the metal surface.
The energy of a photon can be calculated from its wavelength using the equation:
E = hc/λ
where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.
Given that the wavelength of the incident radiation is 350 nm, we have:
f = c/λ = (3.00 x 10^8 m/s) / (350 x 10^-9 m) ≈ 8.57 x 10^14 Hz E = hc/λ = (6.626 x 10^-34 J s) x (3.00 x 10^8 m/s) / (350 x 10^-9 m) ≈ 1.79 eV
Therefore, the maximum kinetic energy of the emitted photoelectrons is:
KEmax = hf - ϕ = (6.626 x 10^-34 J s) x (8.57 x 10^14 Hz) - (3.0 x 1.6 x 10^-19) J ≈ 1.17 eV
a) The minimum frequency required to cause photoemission is equal to the threshold frequency:
fo = 8.9 x 10^14 Hz
Using the same equation as in part 7a), we can calculate the work function:
ϕ = hf0 = (6.626 x 10^-34 J s) x (8.9 x 10^14 Hz) ≈ 5.90 x 10^-19 J = 3.68 eV
b) i. The maximum kinetic energy of the emitted photoelectrons can be calculated using the same equation as in part 7b):
KEmax = hf - ϕ
The energy of a photon with wavelength 250 nm is:
E = hc/λ = (6.626 x 10^-34 J s) x (3.00 x 10^8 m/s) / (250 x 10^-9 m) ≈ 4.97 eV
Therefore, the maximum kinetic energy of the emitted photoelectrons is:
KEmax = hf -
how much does it cost to operate a 120-w lightbulb for 25 min if the cost of electricity is $0.086 per kilowatt-hour?
Answer:
120 W * 25/60 hr = 50 Whr = .05 kwh
.05 kwr * $.086 / kwr =$ .0043 = .43⊄
If a lightbulb of 120-w operate for 25 min then the cost of electricity is $0.086 per kilowatt-hour is $4.3.
What is energy ?Energy is nothing but the ability to do work. there are different energies in different form which are thermal energy, mechanical energy, electric energy and sound energy etc. According to first law of thermodynamic, Energy neither be created nor be destroyed. it can only be transferred from one form into another form. Energy is expressed in joule (J). its dimensions are [M¹ L² T⁻²]. Energy is conserved throughout the motion, according to conservation law of energy, initial energy is equal to final energy.
Power is given by,
P = Work/time.
Given,
Power P = 120 W
Time t = 25 min = 25/60 = 0.416 hr
Total energy needed for the bulb to run 25min is,
Work = Power × time.
Work = 120 W × 0.416 hr.
Work = 50 W-hr
Total cost = 50 W-hr × $0.086 per kilowatt-hour = $4.3
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Jeriah places two metal cubes in contact with each other. Energy and heat flow from Cube A to Cube B. What conclusion can be drawn regarding Cube A?
It is cooler than Cube B.
It is smaller than Cube B.
It is warmer than Cube B.
It is larger than Cube B.
Answer:
C. It is warmer than Cube B.
Explanation:
Option C would be correct because energy as heat flows from warmer objects to cooler objects. If heat flows from Cube A to Cube B, then Cube A must be warmer than Cube B.
Reasoning behind the other option being incorrect:
D: This is incorrect because the direction of heat flow does not depend on the relative size of the cube.
A: This is incorrect because energy as heat would flow from Cube B to Cube A if Cube A is cooler than Cube B.
B: This is incorrect because the direction of heat flow does not depend on the relative sizes of the cubes also.
An energy storage system based on a flywheel (a rotating disk) can store a maximum of 4.9 MJ when the flywheel is rotating at 11000 revolutions per minute. What is the moment of inertia of the flywheel?
The moment of inertia of the flywheel is approximately 0.0337 kg·m^2.
StepsTo solve this problem, we can use the formula for rotational kinetic energy:
KE = (1/2) I ω²
where KE is the kinetic energy of the flywheel, I is its moment of inertia, and ω is its angular velocity.
We can first convert the angular velocity from revolutions per minute (rpm) to radians per second (rad/s):
ω = (11000 rpm) * (2π rad/rev) * (1 min/60 s) = 1146.13 rad/s
Next, we can plug in the values for KE and ω and solve for I:
KE = 4.9 MJ = 4.9 × 10⁶ J
ω = 1146.13 rad/s
(1/2) I ω² = KE
(1/2) I (1146.13 rad/s)² = 4.9 × 10⁶ J
I = (2 * 4.9 × 10⁶J) / (1146.13 rad/s)²
I = 0.0337 kg·m²
Therefore, the moment of inertia of the flywheel is approximately 0.0337 kg·m².
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(9\%) Problem 1: A disk of massMand radiusR, a hoop of mass2Mand radiusR, and a ball of massMand radius2Rare rolling without slipping. The hoop can be treated as a thin ring and the ball should be modeled as a hollow sphere.25%Part (a) The objects are rolling on a flat surface with the same linear speed. Which have the same angular speed? Choose the best answer. Disk and Hoop✓Correct!25%Part (b) The objects are rolling on a flat surface with the same angular speed. Which have the same linear speed? Choose the best answer. Disk and Hoop✓Correct!25%Part (c) Which of the objects has the smallest moment of inertia? Choose the best answer. Disk 、 Correct!▹25%Part (d) The objects are placed at the top of an incline and released from rest. Assuming that the objects roll without slipping, which one is first reach the bottom of the incline? Choose the best answer. \begin{tabular}{llll} \hline Hints: & deduction per hint. Hints remaining: 2 & Feedback: & deduction per feedback. \end{tabular}
Beforehand to hit the bottom will be the disk. The measure of matter contained inside a molecule or item is indicated by its mass, which is represented either by symbol m.
There in International System (SI), the kilogram serves as the default unit of mass (kg). The item moving with the greatest acceleration would descend first. Currently, the formula for any object's acceleration while simply rolling down a slope is
[tex]a = \frac{gsin(theta)}{1 +\frac{1}{MR_{2} } }[/tex]
where is the inclined plane's angle and g is the gravitational acceleration. The body's radius, mass, and inertia time are all represented by the letters M and R, respectively.
speed of a disk is calculated.
Given
1) Mass [tex]= M[/tex]
2) Radius [tex]= R[/tex]
3) [tex]I = \frac{1}{2}MR_{2}[/tex] ⇒ [tex]\frac{1}{MR_{2} } = \frac{1}{2}[/tex]
Hence, disk acceleration [tex]a^{d}[/tex]
[tex]a = \frac{gsin(theta)}{1 + 0.5}[/tex]
[tex]a = \frac{2}{3} gsin (theta)[/tex]
Hoop acceleration calculations
Given:
1) Mass [tex]= 2M[/tex]
2) Radius [tex]= R[/tex]
3) [tex]I = 2MR_{2}[/tex] ⇒ [tex]\frac{1}{2MR_{2} }[/tex] [tex]= 1[/tex]
acceleration [tex]ah[/tex]
[tex]ah = \frac{gsin(theta)}{1 + 1}[/tex]
[tex]ah = \frac{1}{2} gsin (theta)[/tex]
Estimating the ball's acceleration
Given :
1) Mass = [tex]=M[/tex]
2) Radius [tex]= 2R[/tex]
3) [tex]I = \frac{2}{3} M(2R)_{2}[/tex] ⇒ M [tex]\frac{1}{M (2R)2} = \frac{2}{3}[/tex]
acceleration [tex]ab[/tex]
[tex]ab = \frac{gsin(theta)}{1 + \frac{2}{3} }[/tex]
[tex]ab = \frac{3}{5} gsin (theta)[/tex]
By comparing, we obtain [tex]ad[/tex] ≥ [tex]ab[/tex] ≥ [tex]ah[/tex] therefore disk would reach the bottom initially.
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you use the doppler method to discover a planet around a nearby star that is very similar to the sun; the velocity curve that has a period of 6 months. what can you conclude about the planet's orbital distance?
The planet's average distance from the star is approximately 0.78 astronomical units (AU).
What is the planet's orbital distance?Based on the information given, the velocity curve obtained using the Doppler method has a period of 6 months. This means that the planet completes one full orbit around the star in 6 months.
Using Kepler's third law, we can relate the orbital period of the planet to its distance from the star:
(T^2 / a^3) = (4π^2 / GM)
where;
T is the orbital period of the planet, a is its semi-major axis (i.e., its average distance from the star), G is the gravitational constant, andM is the mass of the star.Since we know the orbital period of the planet (6 months) and the mass of the star (similar to the Sun), we can solve for the semi-major axis:
a = (T^2 GM / 4π^2)^(1/3)
Substituting the given values, we get:
a = [(6 months)^2 * (1 solar mass) * (6.67 x 10^-11 N m^2/kg^2) / (4π^2)]^(1/3)
Simplifying the expression, we get:
a ≈ 0.78 AU
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what would be the path of a positive charge moving from left to right through the pair of oppositely charged electric plates in the diagram?
When a positive charge moves from left to right through a pair of oppositely charged electric plates, it follows a specific path due to the electric field produced by the plates.
The electric plates in the diagram have opposite charges. The top plate is positively charged, while the bottom plate is negatively charged. When a positive charge is placed in the electric field, it experiences a force that pushes it towards the negative plate. This is because opposite charges attract each other.
The path that the positive charge follows depends on the strength of the electric field and the speed at which it is moving. If the electric field is weak, the positive charge will not experience a significant force and will move in a straight line from left to right. However, if the electric field is strong, the positive charge will experience a stronger force and will curve towards the negative plate.
The direction of the force can be determined using the right-hand rule. If the positive charge is moving from left to right and the electric field is pointing down (from the positive plate to the negative plate), then the force on the charge will be towards the center of the plates (in the direction of the negative plate).
In summary, the path of a positive charge moving from left to right through a pair of oppositely charged electric plates depends on the strength of the electric field and the speed at which it is moving. If the electric field is weak, the charge will move in a straight line. If the electric field is strong, the charge will curve towards the negative plate. The direction of the force can be determined using the right-hand rule.
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a 1.25 kg hoop with a radius of 11.1 cm rolls without slipping and has a linear speed of 1.50 m/s. find the translational kinetic energy. answer should have two decimal places.
The translational kinetic energy of the hoop is 1.41 kg·[tex]m^2/s^2.[/tex]
To find the translational kinetic energy of the hoop, we will use the following steps:
Step 1: Identify the given values.
The mass (m) of the hoop is 1.25 kg, and the linear speed (v) is 1.50 m/s.
Step 2: Understand the formula for translational kinetic energy.
The formula for translational kinetic energy [tex](K_t)[/tex] is given by:
[tex]K_t = (1/2)mv^2[/tex]
Step 3: Substitute the given values into the formula.
[tex]K_t = (1/2)(1.25 kg)(1.50 m/s)^2[/tex]
Step 4: Calculate the translational kinetic energy.
[tex]K_t = (0.5)(1.25 kg)(2.25 m^2/s^2)[/tex]
[tex]K_t = 1.40625 kg·m^2/s^2[/tex]
Step 5: Round the answer to two decimal places.
[tex]K_t = 1.41 kg·m^2/s^2[/tex]
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Which of the following has a higher kinetic energy?
A 1000 kg car rolling at a speed of 5 m/s across the floor or a 1500 kg hippo swimming at a speed of 2 m/s
the hippo
their kinetic energies are equal
not enough information to determine
the car
Answer:
To solve the problem, we need to use the formula for kinetic energy:
KE = (1/2) * mass * velocity^2
For the car:
KE = (1/2) * 1000 kg * (5 m/s)^2 = 12,500 J
For the hippo:
KE = (1/2) * 1500 kg * (2 m/s)^2 = 3000 J
Therefore, the car has a higher kinetic energy:
Answer: The car.
a 3.1-mm-diameter wire carries a 14 a current when the electric field is 6.9×10−2 v/m .
What is the wire's resistivity?
Wire's resistivity is approximately 3.72 × 10^−8 Ωm.
To find the wire's resistivity, we can follow these steps:
1. Calculate the wire's cross-sectional area.
2. Use Ohm's Law to find the resistance.
3. Use the resistance and cross-sectional area to find the resistivity.
Step 1: Calculate the wire's cross-sectional area.
The wire has a diameter of 3.1 mm, so its radius is 1.55 mm. Convert the radius to meters (1.55 × 10⁻³ m). The cross-sectional area (A) of the wire can be calculated using the formula A = πr², where r is the radius.
A = π(1.55 × 10⁻³ m)² ≈ 7.54 × 10⁻⁶ m²
Step 2: Use Ohm's Law to find the resistance.
Ohm's Law states that V = IR, where V is voltage, I is current, and R is resistance. We know the electric field (E) is 6.9 × 10⁻² V/m and the current (I) is 14 A. To find the voltage (V), we can multiply the electric field by the length of the wire (L). However, we don't know the length of the wire. Instead, we can find the resistance per unit length (R/L) by dividing both sides of Ohm's Law by L:
(V/L) = I(R/L) → E = I(R/L)
Now, we can solve for (R/L):
(R/L) = E/I = (6.9 × 10⁻² V/m) / 14 A ≈ 4.93 × 10⁻³ Ω/m
Step 3: Use the resistance and cross-sectional area to find the resistivity.
Resistivity (ρ) can be calculated using the formula ρ = (R × A) / L. Since we have the resistance per unit length (R/L) and the cross-sectional area (A), we can write the formula as:
ρ = (R/L) × A
Now, plug in the values:
ρ = (4.93 × 10⁻³ Ω/m) × (7.54 × 10⁻⁶ m²) ≈ 3.72 × 10⁻⁸ Ωm
So, the wire's resistivity is approximately 3.72 × 10⁻⁸ Ωm.
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a convex mirror has a focal length of -24 cm. find the magnification produced by the mirror when the object distance is (a)14 cm and (b)16 cm.
The magnification produced by the convex mirror when the object distance is (a) 14 cm is 0.63 and (b) 16 cm is 0.6.To find the magnification produced by a convex mirror with a focal length of -24 cm when the object distance is 14 cm and 16 cm, we'll first need to calculate the image distance for each case using the mirror formula.
Mirror formula = 1/f = 1/v + 1/u, Where f is the focal length, v is the image distance, and u is the object distance.
(a) When the object distance (u) is 14 cm:
1/f = 1/v + 1/u
1/-24 = 1/v + 1/14
Now, solve for the image distance (v):
1/v = 1/-24 - 1/14
1/v = -38/336
v = 336/38
v = -8.84 cm Now we can find the magnification (M) using the formula: M = -v/u
For the object distance of 14 cm:
M = -(-8.84)/14
M = 0.63
(b) When the object distance (u) is 16 cm:
1/f = 1/v + 1/u
1/-24 = 1/v + 1/16
Solve for the image distance (v) again:
1/v = 1/-24 - 1/16
1/v = -40/384
v = 384/40
v = -9.6 cm. we can find the magnification (M) using the formula: M = -v/u
For the object distance of 16 cm:
M = -(-9.6)/16
M = 0.6.
Therefore the magnification produced by the mirror when the object distance is 14 cm is 0.63 and 16 cm is 0.6.
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Which of these best explains the ability of small insects to walk on the surface of still water?
A: Water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid.
B: The insects' feet are coated with ionic compounds.
C: Water has a very high specific heat.
D: Water molecules near the surface produce more buoyant force than water molecules within the liquid.
Answer: water molecules at the surface experience fewer hydrogen bonds than water molecules within the liquid
Explanation: I took the test
"The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6. 0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm? (Assume k = 2. 0 kN/m. )"
The block is moving at 6.0 m/s when the spring is compressed 15 cm.
We can use conservation of energy to solve this problem. The initial kinetic energy of the block will be converted to potential energy when the spring is compressed. At this point, the block will momentarily come to a stop before bouncing back, so its velocity will be zero.
Let's first calculate the potential energy stored in the spring when it is compressed 15 cm:
Δx = 15 cm = 0.15 m (conversion from cm to m)
k = 2.0 kN/m = 2000 N/m (conversion from kN/m to N/m)
x = Δx = 0.15 m
[tex]U = (1/2) k x^2 = (1/2) * 2000 * (0.15)^2 = 22.5 J[/tex]
Now, we can equate this potential energy to the initial kinetic energy of the block:
[tex]K = (1/2) mv^2 = (1/2) * m * (6.0)^2 = 18 m J[/tex]
where m is the mass of the block.
Equating the two equations, we get:
18 m J = 22.5 J
[tex]m = 22.5 J / 18 (m/s)^2 = 1.25 kg[/tex]
Now that we know the mass of the block, we can use conservation of momentum to find its velocity when the spring is compressed:
Before the collision:
[tex]m_1 = 1.25 kg[/tex]
[tex]v_1 = 6.0 m/s[/tex]
After the collision:
[tex]m_2 = 1.25 kg[/tex]
[tex]v_2 = ?[/tex]
Conservation of momentum:
[tex]m_1v_1 = m_2v_2[/tex]
[tex]v_2 = m_1v_1/m_2 = (1.25 kg) * (6.0 m/s) / (1.25 kg) = 6.0 m/s[/tex]
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Before rotating the platform, the hanging mass is disconnected from the test mass and removed from the platform.TrueFalse
The given statement "Before rotating the platform, the hanging mass is disconnected from the test mass and removed from the platform." is True because the concept of a rotating platform involves a disc that rotates about its central axis with a pendulum suspended from its edge.
The plane of rotation and the plane of the pendulum oscillation are separated by a tiny angle. A rotating platform is utilized to generate an artificial gravity environment in space. The centrifugal acceleration produced by rotation is used to imitate the gravitational pull of Earth's mass on objects.
The pendulum is an instrument that measures acceleration, and it functions by oscillating with a period that is dependent on the magnitude of the acceleration it experiences. Besides, it has a number of other applications, including scientific research, engineering tests, and astronaut training in simulated gravity. The test mass is left alone on the platform while the hanging mass is disconnected and removed from the platform before the platform is rotated.
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a farsighted person has a near point of 50 cm . part a what strength lens, in diopters, is needed to bring his near point to 25 cm ?
A farsighted person has a near point of 50 cm, and to bring their near point to 25 cm, we will need a converging lens.
Let the required lens' focal length be f, and the near point be p1. We will now use the lens formula to determine the strength of the lens (in diopters).
Formula: 1/f = 1/p1 + 1/p2
Since the lens formula is expressed in meters, we must first convert the near point to meters: 50 cm = 0.5 mp1 = 0.5 m - 0.25 m (because we want the image to be formed 25 cm away from the eye) = 0.25 m
Putting in these values in the above formula, we get:1/f = 1/0.25 - 1/0.5 1/f = 4 - 2 f = 1/2 f = 0.5 m Diopters are defined as the reciprocal of the focal length in meters, and we get the strength of the lens as:
Strength of the lens = 1/f = 1/0.5 = 2 diopters. Therefore, a lens of 2 diopters strength will be required to bring the farsighted person's near point to 25 cm.
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if you weigh 670 n on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 16.0 km ? take the mass of the sun to be ms
If a person weighs 670 N on Earth, their weight on the surface of a neutron star that has the same mass as our sun and a diameter of 16.0 km would be 9.26 x 10^12 N.
To find out, we'll need to use the formula for surface gravity: g = (GM)/r²
where: g is surface gravity in N/kg, G is gravitational constant in Nm²/kg², M is the mass in kilograms, r is the radius in meters.
First, we'll need to find the mass of the neutron star using the mass of the sun, which is Ms = 1.989 x 10³⁰ kg.
M = Ms = 1.989 x 10³⁰ kg
The radius of the neutron star is 16.0 km or 16,000 m. r = 16,000 m
Now we can plug these values into the surface gravity formula:
g = (GM)/r²g = [(6.674 x 10^-11 Nm²/kg²)(1.989 x 10³⁰ kg)]/(16,000 m)²g = 1.15 x 10^12 N/kg
Finally, to find the weight of the person on the surface of the neutron star, we'll multiply their mass by the surface gravity:
weight = mgweight = (670 N)/(9.81 m/s²)weight = 68.27 kg
weight on neutron star = (68.27 kg)(1.15 x 10^12 N/kg)
weight on neutron star = 9.26 x 10^12 N
Therefore, the weight of the person on the surface of a neutron star that has the same mass as our sun and a diameter of 16.0 km would be 9.26 x 10^12 N.
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In an electromagnetic wave, how are the magnetic field, the electric field, and the direction of propagation oriented to each other?n electromagnetic wave, how are the magnetic field, the electric field, and the direction of propagation oriented to each other а. All three are parallel to each other and are along the x axis. b. All three are mutually perpendicular to each other. c.The electric field and magnetic fields are parallel to each other and perpendicular to the direction of propagation. d. The magnetic field and direction of propagation are parallel to each other along the y axis and perpendicular to the electric field
In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and both are perpendicular to the direction of propagation. This is known as transverse polarization. So, the correct option is (b) "All three are mutually perpendicular to each other."
The electric field and magnetic field are in phase and oscillate perpendicular to each other and to the direction of wave propagation. The direction of propagation is the direction in which the wave travels. The wave can travel in any direction perpendicular to the fields. The speed of the electromagnetic wave is determined by the properties of the medium in which the wave is traveling and is given by the equation v = c/n, where c is the speed of light in vacuum and n is the refractive index of the medium.
Electromagnetic waves are a form of energy that can travel through a vacuum and do not require a medium for their propagation. Examples of electromagnetic waves include radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays. Hence, option b is correct.
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If a 50.0-lb tire sample is dragged though a parking lot with a force of 8.0 lbs. , determine the friction coefficient between the tire and the pavement.
A. 0.20
B. 0.16
C. 0.35
D.6.25
Show the work for determining the friction of the tire...show symbolic solution then numerical solution.
To determine the friction coefficient between the tire and the pavement, we need to use the equation for friction. Therefore, the correct answer is option B: 0.16.
Using an example, what is friction force?a) The force produced between surfaces that slide against one another is referred to as the frictional force. b) Since frictional force develops when two surfaces come into contact with one another, it is referred to be a contact force. Walking on the road is an illustration of frictional force.
Friction force = friction coefficient x normal force
where the normal force is the force perpendicular to the surface, which in this case is the weight of the tire (50.0 lb). The friction force is the force required to drag the tire through the parking lot, which is 8.0 lb.
Substituting the given values, we get:
8.0 lb = friction coefficient x 50.0 lb
Solving for the friction coefficient, we get:
friction coefficient = 8.0 lb / 50.0 lb = 0.16
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how coulomb's law is the scientific concept explaining the intraparticle relationship between electrons and protons:
Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
In the context of atoms, electrons (negatively charged) and protons (positively charged) experience an attractive force, which helps maintain the stability of the atom. The Coulomb force, also known as Coulomb's law, is an essential concept in physics. Coulomb's law describes the force between two charged particles based on their charges and the distance between them. Coulomb's law of electrostatics is a fundamental law in physics that describes the relationship between electrically charged particles.
Coulomb's law is a simple equation that states that the force between two charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. In other words, Coulomb's law can be used to determine the force of attraction or repulsion between charged particles, such as electrons and protons, based on their charges and separation distance.
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The movement of crustal plates results from circulating currents in material beneath the crust of the Earth. Which best describes the material which moves the crustal plates?
- hot water
- molten rock
- liquid metal
- solid iron
The correct option is B, The movement of crustal plates results from circulating currents in material beneath the crust of Earth. Molten rock best describes the material which moves the crustal plates.
Molten rock, also known as magma, is a hot, fluid material that exists beneath the Earth's surface. It is composed of a mixture of melted rock, gases, and minerals. Magma is formed when the Earth's mantle or crust melts due to heat and pressure, or when the mantle releases gases that cause rocks to melt.
Magma can have different compositions depending on the type of rock it originated from. For example, basaltic magma is composed of dark, dense rocks and has a low viscosity, which means it flows easily. Andesitic magma, on the other hand, is composed of lighter, more viscous rocks and is less fluid. Lava can flow or explode out of volcanoes, creating new landforms and changing the landscape.
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Complete Question: -
The movement of crustal plates results from circulating currents in material beneath the crust of Earth. Which best describes the material which moves the crustal plates?
a. hot water
b. molten rock
c. liquid metal
d. solid iron
The smith family is traveling in their car at 50 km/h due east. Mr. Smith is using cruise control to maintain a constant speed. Describe the net force acting on the Smith car.
A. Net force equals zero.
B. Net forces are unbalanced
C. There is no way to determine net force
D. Net force is positive and to the east
A. Net force equals zero. The net force acting on the Smith car is equal to zero.
The net force acting on a car on cruise control at constant speed.This is because the car is traveling in a straight line at a constant speed, meaning that the forces acting on the car are balanced. The car is being propelled forward by an applied force, such as the engine, and this force is counteracted by the force of friction from the ground. The net force is the sum of all these forces, and since they are all balanced, the net force is also equal to zero.
Cruise control helps maintain a constant speed, as it adjusts the engine power to counteract the changing frictional forces on the car. This ensures that the forces on the car remain in balance, so the net force remains equal to zero.
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Consider the mass and velocity values of Object A and B below. Compared to Object B, Object A has _______ momentum.
Object A: m=67 kg v=17m/s
Object B: m=2 kg v=100m/s
The momentum of object B is higher than that of object A.
Is momentum greater the higher the mass, velocity, or both?A system's mass and velocity are multiplied to produce its linear momentum. It is simple to see how momentum relates to an object's mass and speed. As a result, an object's momentum grows as its mass or speed increases.
Mass times speed equals momentum.
Object A: momentum is equal to 67 kg times 17 m/s, or 1139 kg/m/s.
Object B's momentum is equal to 2 kg times 100 m/s, or 200 kg/m.
Given that momentum is directly inversely proportional to both mass and velocity, object B has more momentum than object A since its lower mass is more than offset by its much higher velocity.
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the higher the r-value, the greater the insulating effectiveness. (1 point) group of answer choices true false
It is true that the higher the R-value, the greater the insulating effectiveness of a material.
True.
The R-value is a measure of a material's insulating effectiveness.
A higher R-value indicates greater insulation effectiveness, as it represents the material's resistance to heat flow.
This means that a material with a high R-value will be more effective at insulating and maintaining temperature differences between the interior and exterior environments.
To provide a concise explanation:
1. R-value measures a material's insulating effectiveness.
2. A higher R-value indicates greater resistance to heat flow.
3. Materials with high R-values are better at insulating and maintaining temperature differences.
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a hair dryer draws 1 350 with, a curling iron draws 700 w, and an electric light fixture draws 550 w. if all three of these appliances are operating in parallel on a 120-v circuit, what is the total current drawn?
The total current drawn in the circuit was calculated to be 21.7 A.
Given the appliances are in parallel So the voltage across all the appliances is the same i.e. V = 120 V
We know that P = IΔV
The current in the hair dryer is
I₁ = 1350/120
I₁ = 11.25 A
The current in the curling iron
I₂ = 700/120
I₂ = 5.83 A
The current inside the electric light fixture
I₃ = 550/120
I₃= 4.58 A
So the total current drawn (I) is equal to
I = I₁+ I₂ +I₃
I = 11.25 A +5.83 A +4.58 A
I = 21.7 A
Since all components in a parallel circuit have the same electrical junctions, the voltage across parallel components is the same. The total current is equal to the sum of each individual branch current.
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a 0.400 kg mass hangs from a string with a length of 0.9 m, forming a conical pendulum. the period of the pendulum in a perfect circle is 1.4 s. What is the tension on the string?
The string is under 4.82N of tension.
Tension is the pulling force exerted on a string, rope, cable, or wire when it is stretched or pulled. The tension on a string is equal to the amount of force applied to it divided by its cross-sectional area.
The tension on the string of a conical pendulum can be calculated using the equation:
[tex]T =\frac{ (4\pi^2m)}{(L^2T^2)},[/tex]
where T is the string's tension, m is the pendulum's mass, L is the string's length, and T is the pendulum's period of motion.
m=mass of pendulum=0.400kg
L=length of string =0.9m
T=pendulum's period of motion=1.4s
Plugging in the given values, we get:
[tex]T =\frac{ (4 \pi^2 * 0.400 kg) }{ (0.9 m^2 * 1.4 s^2)}\\\\T = 4.82 N[/tex]
Therefore,The Tension on the string is 4.82N
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diana raises a 1000 n piano a distance of 5.00 m using a set of pulleys. she pulls in 20.0 m of rope. if the actual force is 300 n, what is the actual mechanical advantage?
The actual mechanical advantage is 6.67 .
Mechanical advantage is the ratio of the output force to the input force in a machine. It is a ratio that specifies the multiple by which the input force is increased to produce the output force.
MA = Output Force / Input Force
Now, let's solve the given problem:
Input Force = 300 N
Output Force = ? , MA = ? , MA = Output Force / Input Force
Output Force = MA × Input Force
Output Force = (1000 N / 300 N) × Input Force
Output Force = 3.33 × Input Force
Diana pulled in 20 m of rope, thus the rope multiplied her force.
Therefore, the distance moved by the rope is the input distance, and the distance moved by the piano is the output distance.
Output Distance / Input Distance = MA
Output Distance = 5 m
Input Distance = 20 m
MA = Output Distance / Input Distance
MA = 5 m / 20 m
MA = 0.25MA = 1 / 0.25MA = 4
Output Force = MA × Input Force
Output Force = 4 ×300 N
Output Force = 1200 N
The actual mechanical advantage is equal to the output force divided by the input force. This is also equal to the number of times the machine increases the force applied to it.
So, Actual mechanical advantage = Actual output force / Actual input force = 300 N / (1200 N / 3.33)Actual mechanical advantage = 6.67 .Hence, the actual mechanical advantage is 6.67.
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a 25.0kg child jumps to the ground from a structure 1.00m high and comes to rest 0.500s after first contact with the ground. what average force is applied by the child in coming to rest
Answer:
Approximately [tex]470\; {\rm N}[/tex].
(Assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex] and that air resistance is negligible.)
Explanation:
Under the assumptions, the acceleration of the child would be constantly [tex]a = g = 9.81\; {\rm N \cdot kg^{-1}} = 9.81\; {\rm m\cdot s^{-2}}[/tex] while the child was in the air.
Apply the SUVAT equation [tex]v^{2} - u^{2} = 2\, a\, x[/tex] to find the velocity [tex]v[/tex] of the child right before landing:
[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x}\end{aligned}[/tex], where:
[tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] is the initial velocity of the child,[tex]a = 9.81\; {\rm m\cdot s^{-2}}[/tex] is the vertical acceleration, and[tex]x = 1.00\; {\rm m}[/tex] is the vertical displacement (change in height.)The child is at rest [tex]\Delta t = 0.500\; {\rm s}[/tex] after contact. During that [tex]0.500\; {\rm s}[/tex], velocity would have changed by [tex]\Delta v = \sqrt{u^{2} + 2\, a\, x}[/tex]. Momentum of the child would have changed by [tex]\Delta p = m\, \Delta v[/tex], where [tex]m = 25.0\; {\rm kg}[/tex] is the mass of the child.
Divide this change in momentum by the duration [tex]\Delta t[/tex] to find the average net force:
[tex]\displaystyle F_{\text{net}} = \frac{m\, \Delta v}{\Delta t}[/tex].
There are two forces on the child: upward normal force from the ground [tex]F_{\text{normal}}[/tex] and downward gravitational attraction [tex]m\, g[/tex] from the Earth. The resultant force on the child points upwards:
[tex]F_{\text{net}} = F_{\text{normal}} - m\, g[/tex].
Rearrange this equation to find the normal force on the child:
[tex]\begin{aligned}F_{\text{normal}} &= F_{\text{net}} + m\, g \\ &= \frac{m\, \Delta v}{\Delta t} + m\, g \\ &= \frac{m\, \sqrt{2\, a\, x + u^{2}}}{\Delta t} + m\, g\\ &= \frac{(25.0)\, \sqrt{2\, (9.81)\, (1.00) + 0^{2}}}{0.500}\; {\rm N} + (25.0)\, (9.81)\; {\rm N}\\ &\approx 470\; {\rm N}\end{aligned}[/tex].
This normal force from the ground on the child is the reaction to the force that the child exerted on the ground. The two forces will have the same magnitude: approximately [tex]470\; {\rm N}[/tex]. Hence, the child would have exerted an average force of approximately [tex]470\; {\rm N}\![/tex] on the ground during that [tex]0.500\; {\rm s}[/tex].