Answers
The charge on the segment ds is [tex]dQ=\frac{Q}{\pi a}ds=\frac{Q}{\pi a}ad\theta= \frac{Q}{\pi} d\theta[/tex]
As Q is uniformly distributed over semicircel so,charge density
[tex]\lambda=\frac{Q}{s}=\frac{Q}{\pi r}[/tex]
so in ds region charge: [tex]dQ=\frac{Q}{\pi a}ds=\frac{Q}{\pi a}ad\theta= \frac{Q}{\pi} d\theta[/tex]
In physics, a uniformly distributed charge refers to a charge distribution where the charge density is constant throughout the given volume or surface. In other words, the amount of charge per unit volume or unit area is the same everywhere within the region.
Uniformly distributed charge is an important concept in electrostatics, which is the study of the behavior of electric charges at rest. The electric field produced by a uniformly charged object has a particularly simple form, and this makes it easier to calculate the electric field at any point outside the charged object using Gauss's Law.
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the complete question is:
What is the charge on a segment ds?
Express your answer in terms of some, all, or none of the variables Q, a, θ, dθ, and the constant π.
Related Questions
In the figure, a small block of mass m = 0.019 kg can slide along the frictionless loop-the-loop, with loop radius R = 13 cm. The block is
released from rest at point P, at height h = 5R above the bottom of the loop. How much work does the gravitational force do on the
block as the block travels from point P to (a) point Q and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point P, (d) at point Q, and (e) at the top of the loop?
Answers
The amount of work done as the block travels from point P to (a) point Q is 0.096J, (b) the top of the loop is 0.072J, the potential energy when the block is (c) at point P is 0.121 J, (d) at Q is 0.024 J, (e) at top is 0.048 J
Given the mass of small block (m) = 0.019kg
The radius of loop (R) = 13cm.
The height of point P (H) = 5R = 5 *13 = 65cm
the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop.
We know that the potential energy is calculated as (PE) = m*g*h where g is the gravitational acceleration = [tex]9.8m/s^2[/tex]
(a) The work done by the gravitational force on the block as the block travels from point P to point Q is: W = F * d = m*g*d
the distance from point P to Q = 4R = 4 * 13 = 52cm
W =[tex](0.019 kg)*(9.8 m/s^2)*(4R) = 0.096J[/tex]
(b) The work done by the gravitational force on the block as the block travels from point P to the top of the loop is:
Here the displacement from point P to top of loop = H - 2R = 3R = 39cm
W = m*g*h = [tex](0.019 kg)*(9.8 m/s^2)*(3R) = 0.072J[/tex]
(c) The gravitational potential energy of the block-Earth system at point P is: PE = mgh = [tex](0.019 kg) * (9.8 m/s^2) * (5R) = 0.121 J[/tex]
(d) The gravitational potential energy of the block-Earth system at point Q is: PE = mgh = [tex](0.019 kg) * (9.8 m/s^2) * (R) = 0.024 J[/tex]
(e) The gravitational potential energy of the block-Earth system at the top of the loop is:
PE = mgh =[tex](0.019 kg)*(9.8 m/s^2)*(2R) = 0.048 J[/tex]
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Q) A velocity field is given as V = (ay,−ax + abt,0 ),
a.) Find the streamline equation for this flow field.
b.) Plot at least 3-streamlines in the xy-plane for a=1, and b=1.
c.) Indicate the direction of the flow on each streamline at point (2,3) in the
first quadrant.
Answers
For a vector field:
a) the streamline equation is y = Ce^(-0.5ax²+abty)
b) three streamlines include; y = e^(-0.5x²+bt), y = 2e^(-0.5x²+bt), y = 3e^(-0.5x²+bt)
c) direction of flow at points (2,3) is (3a, -2a+3bt)
How to calculate streamlines?a) To find the streamline equation, use the definition that the velocity vector is tangent to the streamline at every point along the streamline. Let (x,y) be a point on a streamline, then:
dx/dt = a y
dy/dt = -a x + abt
Using the chain rule:
dy/dx = (dy/dt)/(dx/dt) = (-a x + abt)/(a y)
Integrating both sides:
ln |y| = -0.5 a x² + abt y + C
where C is a constant of integration. Solving for y:
y = Ce^(-0.5ax²+abty)
This is the equation of a streamline.
b) To plot the streamlines, use the equation we derived in part (a) and choose different values of C to get different streamlines. For example, if we choose C = 1, 2, 3, then the streamlines will be:
y = e^(-0.5x²+bt), y = 2e^(-0.5x²+bt), y = 3e^(-0.5x²+bt)
c) To indicate the direction of the flow on each streamline at point (2,3), we need to evaluate the velocity vector at that point.
V = (ay,-ax + abt,0)
At point (2,3):
V = (3a, -2a+3bt, 0)
The direction of the flow is given by the direction of the velocity vector. In this case, the velocity vector points in the direction of (3a, -2a+3bt), which depends on the values of a, b, and t.
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A 5.0KG block is placed at rest on a 30 degree incline. The coefficient of static friction is 0.5 and the coefficient of kinetic friction is lower than that. Draw a free body diagram of the block including all force components and determine the net force. Will the block slide or will it remain at rest?
Answers
Here is a free body diagram of the block on the incline:
/|
/ |
/ |
/ |
/ |
/ |
/θ |
/ |
/ |
/ |
/____________|
N
|
|
|
|<----f_kinetic
|_______ <----f_gravity
Static force:
f_static is less than the maximum force of static friction, the block will not slide and will remain at rest.
In this diagram, θ represents the angle of the incline, N represents the normal force exerted on the block by the incline, f_gravity represents the force of gravity pulling the block down the incline, and f_kinetic represents the force of kinetic friction opposing the motion of the block. Since the block is at rest, the net force must be zero.
We can calculate the force of gravity using the formula:
f_gravity = mgcosθ
where m is the mass of the block, g is the acceleration due to gravity (9.81 m/s^2), and θ is the angle of the incline. Substituting the given values, we get:
f_gravity = (5.0 kg)*(9.81 m/s²)*cos(30°) = 42.7 N
The normal force, N, is perpendicular to the incline and equal in magnitude to the component of the force of gravity perpendicular to the incline. We can calculate it using the formula:
N = f_gravity*sinθ
Substituting the given values, we get:
N = (5.0 kg)*(9.81 m/s²)*sin(30°) = 24.5 N
The force of static friction, f_static, can be found using the formula:
f_static = μ_static*N
where μ_static is the coefficient of static friction. Substituting the given value, we get:
f_static = (0.5)*(24.5 N) = 12.3 N
Since the block is at rest, the force of static friction must be equal in magnitude to the component of the force of gravity parallel to the incline:
f_static = f_gravitysinθ = (5.0 kg)(9.81 m/s²)*sin(30°) = 24.5 N
Since f_static is less than the maximum force of static friction, the block will not slide and will remain at rest.
We don't need to determine the force of kinetic friction since the block is not sliding.
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Two particles are initially separated by 20 cm. Particle A, with mA = 10pg and qA = -5 nC, is on the left and makes a head-on collision with Particle B at a speed of 4 * 10^4 m/s. Particle B, with mB = 5pg and qB = -10nC, is on the right and moves toward Particle A at a speed of 6 * 10^4 m/s. Assume Particle A is moving in the positive direction. (If you wish to get correct answers, then be sure to use ke = 9 * 10^9 N * m^2 / C^2 and e = 1.6 * 10^-19 C.)
(a) The initial mechanical energy of the system in muJ (μJ - microjoules) and to three decimal places, is?
(b) The magnitude of the maximum force acting on qA during the collision in mN (miliNewton) and to three decimal places, is?
(c) The work done by the electric force on the system to stop the particles in muJ (μJ) and to two decimal places, is?
(d) The minimum separation distance between the two particles in cm and to three decimal places, is?
(e) The maximum speed experienced by Particle A is
Option 1: when it is infinitely far away from Particle B
2: is unable to be determined
3: when it is closest to Particle B
4: at its initial location
5: at a location that cannot be determined without more information
Answers
When particle A is infinitely far from particle B, option 1, it moves at its top speed.
Initial velocity of the first particle was u₁ = 4×10 ⁴m/s Initial velocity of the second particle was u₂ = 6×10⁴ m/s Initial velocity of the third particle was v = Final velocity of both particles after collision was v =
Use the idea of linear momentum conservation;
a) After a collision, a 5 kg particle moves at a speed of -1 m/s, changing the total kinetic energy of the system by -40 joules. More kinetic energy can be obtained by adding an external substance.k e = 9 ₓ 10⁹ N m² / C² and e = 1.6 ₓ 10⁻¹⁹ C.)
b) The Joule is the unit used to measure kinetic energy, which is the energy that an item stores as a result of motion. The product of a particle's mass and velocity is known as momentum. In order to determine the particle's velocity following a collision, we must first compare the values in the momentum conservation formula, which is provided as:c) m1 and v1 represent the mass and speed of a 10 pg item (before collision)m2, v2 represent the mass and speed of a 10 kilogram item (before collision)
M1, V1 stand for mass and speed.
d) A first particle and a second particle are O.v(m₁+ m₂) = m₁u₁ - m₂u₂.
11 x 2 - 11 x 2 = v( 11 + 11)
22 - 22 = v(22) (22)
0 = 22v = 0/22 = 0
e) After colliding, both particles' final velocities are zeroAs a result, both particles are at rest.
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Can someone please help me find the equivalent resistance of this circuit.
Answers
The equivalent resistance of 5 resistors in parallel, each having a value of "R" is R/5.
What is the equivalent resistance of parallel circuit?
A parallel circuit is an electrical circuit that has two or more paths for the current to flow through. In a parallel circuit, the components are connected in such a way that the voltage across each component is the same, while the current through each component may be different
The given circuit has 5 resistors arranged parallel to each other.
The equivalent resistance of 5 resistors in parallel, each having a value of "R" is given by:
1/R_eq = 1/R + 1/R + 1/R + 1/R + 1/R
Simplifying the equation, we get:
1/R_eq = 5/R
Multiplying both sides by R, we get:
R_eq = R/5
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A basketball has a mass of approximately 624 grams [g] and a volume of 0.25 cubic feet [ft3]. Determine the density of the basketball in units of slug per gallon [slug/gal].
Answers
The basketball has a slugs density of 0.0228 per gallon (m/V = 0.0427 slugs/1.8697 gallons) (approximately).
What in density is a slug?Based on normal gravity, the international foot, and the avoirdupois pound, one slug weighs 32.1740 lb (14.59390 kg). An item with a mass of 1 slug would weigh around 32.2 lbf, or 143 N, at the Earth's surface.
We need to apply the following conversion factor to convert grammes to slugs:
1 slug = 14.5939 kg
1 kg = 1000 g
Therefore:
1 slug = 14.5939 kg = 14.5939 x 1000 g = 14593.9 g
So, the mass of the basketball in slugs is:
m = 624 g / 14593.9 g/slug = 0.0427 slugs
To convert from cubic feet to gallons, we need to use the following conversion factor:
1 gallon = 0.133681 ft^3
Therefore:
0.25 ft⁻³ = 0.25 / 0.133681 = 1.8697 gallons
So, the volume of the basketball in gallons is:
V = 1.8697 gallons
The density of the basketball in slug/gal is:
ρ = m / V = 0.0427 slugs / 1.8697 gallons = 0.0228 slug/gal (approximately)
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HELP NEEDED!!!!!!! [ Reward for answering *Brianliest* ]
Answers
Answer:
Amps are units of current.
Volts are units of voltage.
Ohms are units of resistance.
Explanation:
You're welcome.
A small glass ball is rubbed with a piece of silk, giving the ball a charge of 1.0 x 10^-8 C. Determine the magnitude of the force due to the Earth's magnetic field if the ball is thrown to the west with a velocity 8.0 m/s . The earth's magnetic field is 5.0 x 10^-5 T
Answers
The magnitude of the force due to the electric field on the charged glass ball is 2.4 x [tex]10^{-5}[/tex] N, and the Earth's magnetic field is not relevant in this scenario.
What is Magnetic Field?
A magnetic field is a field created by a magnet, moving electric charge, or changing electric field. A magnetic field can also be created by a loop of electric current. A magnetic field is a vector field, meaning it has both magnitude and direction.
The Earth's magnetic field is not relevant for the interaction between the charged glass ball and the Earth's gravitational field. Instead, we need to calculate the force due to the electric field generated by the charge on the ball.
We can use the formula for the electric force on a charged particle:
F = qE
where F is the force on the charge q, and E is the electric field at the location of the charge.
In this case, the charge on the ball is q = 1.0 x 10^-8 C, and the velocity of the ball is directed to the west, so the direction of the force should be to the north or south.
Assuming the electric field due to the charge on the ball is uniform and perpendicular to the velocity of the ball, we can use the formula for the electric field due to a point charge:
E = k*q / [tex]r^{2}[/tex]
where k is Coulomb's constant (9 x 10^9 N·[tex]m^{2}[/tex]/[tex]C^{2}[/tex]), q is the charge on the ball, and r is the distance from the charge to the point where we want to calculate the electric field.
If we assume that the ball is moving at a constant height above the Earth's surface, then the distance r is constant and we can use the above equation to find the electric field E.
E = k*q / [tex]r^{2}[/tex] = (9 x 10^9 N·[tex]m^{2}[/tex] /[tex]C^{2}[/tex]) * (1.0 x 10^-8 C) /[tex]r^{2}[/tex]
We don't know the distance r, but we do know that the electric force on the ball due to this field must be equal to the force required to cause the ball to move in a circular path, as it is in this case. The force required to maintain a circular motion of radius r with speed v is
where m is the mass of the ball. This force must be equal to the electric force on the ball:
F = qE
We can equate these two expressions to solve for the distance r
Plugging in the given values, we get:
r = 1.78 x [tex]10^{-3}[/tex] m
So the ball is moving in a circular path with radius r = 1.78 x [tex]10^{-3}[/tex]m, and the electric field at the location of the ball is:
E = 2.4 x [tex]10^{3}[/tex] N/C
Finally, we can calculate the force on the charged ball due to this electric field:
F = qE = (1.0 x [tex]10^{8}[/tex] C) * (2.4 x [tex]10^{3}[/tex] N/C)
F = 2.4 x [tex]10^{-5}[/tex]N
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Two point charges q1 and q2 are arranged in a vertical straight line, as shown. Point A is located halfway between q1 and q2.
The point charges are q1 = 5 μC and q2 = -10 μC.
What is the magnitude of the net electric field at point A ?
A: −2.81 × 108 N/C
B: −1.12 × 108 N/C
C: 8.43 × 107 N/C
D: 3.47 × 108 N/C
Answers
The net electric field at point A is -1.12 108 N/C in size.
What do charges Q1 Q2 0 and Q1 Q2 0 mean in terms of electric charge?In light of the fact that both the charge q1 and the other charge q2 are equal to zero. According to the equation, one charge is positive and the other is negative. Both charges are of similar size. This indicates that the two supplied charges on the system will add up to a total charge of zero.
E = k*q/r²
where E is the electric field, q is the charge, r is the distance from the charge, and k is Coulomb's constant, which has a value of 8.99 × 10^9 N·m²/C².
d1 = d2 = (1/2) * (0.1 m) = 0.05 m
E1 = k*q1/d1² = (8.99 × 10⁹ N·m²/C²) * (5 × 10⁻⁶ C) / (0.05 m)² = 1.8 × 10⁸ N/C
E2 = k*q2/d2² = (8.99 × 10^⁹ N·m²/C²) * (10 × 10⁻⁶ C) / (0.05 m)² = 3.6 × 10⁸ N/C
E net = E1 - E2 = 1.8 × 10⁸ N/C - 3.6 × 10⁸ N/C = -1.12 108 N/C.
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a block of wood ispushed against a relaxed spring to compressed it 0.080m the spring constant of spring is 2000 N/m what is true about change in kinetic potential energy of the block spring system
Answers
Change in kinetic energy is 6.4 Joules when a block of wood is pushed against a relaxed spring to compressed it 0.080m.
What is spring constant ?The spring constant is the force necessary to stretch or compress a spring divided by the spring's length. It is used to determine the stability or instability of a spring and, by extension, the system for which it is designed. The symbol k stands for the "spring constant," a value that indicates how "stiff" a spring is. If k is large, it signifies that stretching it a specific length requires more force than stretching a less stiff spring the same length.
Use formula
change in kinetic energy = [tex]\frac{1}{2}[/tex] × k × [tex]x^{2}[/tex]
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1 Fig B U 99 m 33 m A student stands at P so that his distance from building A is 33 m. After clapping his hands once, he hears several echoes. The speed of sound in air is 330 m/s. a) Calculate the time interval between clapping his hands and hearing (i) the first echo, (ii) the third echo chamber containing air.
Answers
Using the principle of the echo, we can see that the first echo can be heard after 0.2 s.
What is echo?In acoustics, an echo refers to the reflection of sound waves off a surface and back to the listener. When sound waves encounter a hard surface, such as a wall or mountain, some of the energy is reflected back towards the source.
To find the time interval for the first echo;
v = 2x/t
t = 2x/v
t = 2(33)/330
t = 0.2 s
Echoes are often heard in large open spaces such as auditoriums, canyons, or empty rooms. They can also be artificially created using electronic sound processing equipment, such as reverb effects in music production or sound systems in large events.
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On a 2,000 km trip, you stop once for gas, once at a rest stop and once for lunch. What is
your average speed if it takes you 10 hours to get to your destination?
Answers
The journey is completed at an average speed of 285.7 km/h.
How fast is 10 kilometres in 30 minutes?10 kilometres is the distance. Time is measured in 30 minute increments. We are aware that time*speed equals distance. Thus, the formula for speed is: speed = distance/time (10/0.5) = 100/5 = 20 km/h. That is speed = distance time. Alternatively, you can calculate the time by dividing the distance travelled by the speed.
We may divide the total distance travelled by the amount of time spent driving to determine the average speed:
Average speed = total distance/time spent driving
Total distance = 2,000 km
Time spent driving = 7 hours
Average speed = 2,000 km / 7 hours
Average speed = 285.7 km/hour
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An average froghopper insect has a mass of 12.8 mg and jumps to a maximum height of 293 mm when its takeoff angle is 62.0∘ above the horizontal. With the takeoff speed being 2.71 m/s :
a) How much kinetic energy did the froghopper generate for this jump? Express your answer in microjoules.
b) How much energy per unit body mass was required for the jump? Express your answer in joules per kilogram of body mass.
Answers
Answer:
Explanation:
a) The potential energy gained by the froghopper at the maximum height of 293 mm can be calculated using the formula:
ΔPE = mgh
where ΔPE is the change in potential energy, m is the mass, g is the acceleration due to gravity, and h is the maximum height.
Substituting the given values, we get:
ΔPE = (12.8 × 10^-6 kg) × (9.81 m/s^2) × (0.293 m) = 3.69 × 10^-6 J
The kinetic energy of the froghopper at takeoff can be calculated using the formula:
KE = 0.5mv^2
where KE is the kinetic energy and v is the takeoff speed.
Substituting the given values, we get:
KE = 0.5 × (12.8 × 10^-6 kg) × (2.71 m/s)^2 = 4.75 × 10^-5 J
Therefore, the total energy generated by the froghopper for the jump is the sum of the potential and kinetic energy, which is:
Total energy = ΔPE + KE = 3.69 × 10^-6 J + 4.75 × 10^-5 J = 5.12 × 10^-5 J
Expressing the answer in microjoules, we get:
Total energy = 5.12 × 10^-5 J = 51.2 µJ
b) The energy per unit body mass required for the jump can be calculated by dividing the total energy generated by the froghopper by its body mass.
Substituting the given values, we get:
Energy per unit body mass = (5.12 × 10^-5 J) ÷ (12.8 × 10^-6 kg) = 4 J/kg
Therefore, the energy per unit body mass required for the jump is 4 J/kg.
A spring stretches 6.0 cm when a 0.25 kg block is hung from it.
If a 0.80 kg block replaces the 0.25 kg block, how far does the spring stretch?
Answers
When a 0.80 Kg block is used in place of a 0.25 Kg block, the length of the string is increased by 19.6 cm.
How does spring stretch become calculated?The formula used by the Hooke's Law Calculator is Fs = -kx, where F is the spring's restoring force, k is the spring constant, and x is the displacement, or the length by which the spring is being stretched
We'll start by determining the spring's string constant. Specifics below:
Extension (e) = 6.0 cm
Mass (m) = 0.25 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) = mg = 0.25 × 9.8 = 2.45 N
Spring constant (K) =?
F = Ke
2.45 = K × 6
Divide both sides by 6
K = 2.45 / 6
K = 0.40 N/cm
The extension will be calculated after the 0.25 kg block is replaced with the 0.80 kg block. As demonstrated below:
Mass (m) = 0.80 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) = mg = 0.80 × 9.8 = 7.84 N
Spring constant (K) = 0.40 N/cm
Extension (e) = ?
F = Ke
7.84 = 0.40 × e
Divide both sides by 0.40
e = 7.84 / 0.40
e = 19.6 cm
We may thus deduce from the preceding computation that the spring will lengthen by 19.6 cm.
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PLEAS HELP ME WITH THIS WORKSHEET PLEASEEEEE!!!!!
Explosion
1) Two swimmers are floating on a raft that is motionless. One swimmer has a mass of 50 kg and
the other at 80 kg. They both push off the raft at the same time. The 80 kg swimmer moves
away at 3 m/s. What velocity does the 50 kg swimmer move away with?
M1 = 50 kg v1' =____ M2 = 80 kg v2' = 3 m/s
Equation: 0= m1 (v1') + m2 (v2')
Elastic
2) Two hockey players are skating towards each other. A 90 kg player traveling at 6 m/s
rams into a 60 kg player moving at 2 m/s. After the collision, the 90 kg player slows to 4
m/s but is still traveling in the same direction. What is the velocity of the 60 kg player?
Equation: m1 (v1) + m2 (v2) = m1 (v1') + m2 (v2')
v2 = -2 m/s
M1 = 90 kg
v1 = 6 m/s M2 = 60 kg
V1' = 4 m/s
v2' =___
Answers
We can use the conservation of momentum to solve both problems:
Conservation of momentum:
0 = m1(v1') + m2(v2')
where m1 = 50 kg, v2' = 3 m/s, and m2 = 80 kg. We can solve for v1' to get:
v1' = -(m2/m1) v2'
v1' = -(80 kg/50 kg) (3 m/s) = -4.8 m/s
Therefore, the 50 kg swimmer moves away from the raft with a velocity of -4.8 m/s.
Conservation of momentum:
m1(v1) + m2(v2) = m1(v1') + m2(v2')
where m1 = 90 kg, v1 = 6 m/s, m2 = 60 kg, and v1' = 4 m/s. We can solve for v2 to get:
v2 = (m1v1 + m2v2 - m1v1') / m2
v2 = (90 kg)(6 m/s) + (60 kg)(2 m/s) - (90 kg)(4 m/s) / 60 kg
v2 = -1 m/s
Therefore, the velocity of the 60 kg player after the collision is -1 m/s, which means they are moving in the opposite direction to the 90 kg player.
The half life of a radioactive element is 4×10⁸ years. Calculate its decay constant and mean life
Answers
The decay constant (λ) of a radioactive element can be calculated using the formula:
λ = ln(2) / T1/2
where ln(2) is the natural logarithm of 2 and T1/2 is the half-life of the element.
Substituting the given values, we get:
λ = ln(2) / (4 x 10^8)
λ = 1.73 x 10^-9 per year
Therefore, the decay constant of the radioactive element is 1.73 x 10^-9 per year.
The mean life (τ) of a radioactive element can be calculated using the formula:
τ = 1 / λ
Substituting the calculated value of λ, we get:
τ = 1 / (1.73 x 10^-9)
τ = 5.78 x 10^8 years
Therefore, the mean life of the radioactive element is 5.78 x 10^8 years.
a spanner is dropped from a sixth floor window and take 2.2s to hit the ground. A)calculate the height from wich it was drop.b)its impact velocity
Answers
a) The spanner was dropped from a height of 24.2 meters. b)The impact velocity of the spanner is 21.6 meters per second.
What exactly are velocity and example?Simply put, velocity is the rate at which something travels in a specific direction.
We can use the equations of motion to solve this problem.
a) To calculate the height from which the spanner was dropped, we can use the equation:
[tex]h = (1/2)gt^2[/tex]
where h is the height, g is the acceleration due to gravity, and t is the time taken to fall.
Substituting the given values, we get:
[tex]h = (1/2) \times 9.81 m/s^2 \times (2.2 s)^2 \\= 24.2 meters[/tex]
Therefore, the spanner was dropped from a height of 24.2 meters.
b) To calculate the impact velocity of the spanner, we can use the equation:
v = gt
where v is the final velocity, g is the acceleration due to gravity, and t is the time taken to fall.
Substituting the given values, we get:
[tex]v = 9.81 m/s^2 \times 2.2 s \\= 21.6 m/s[/tex]
Therefore, the impact velocity of the spanner is 21.6 meters per second.
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EARTH AND SPACE SCIENCE!
All of the following are Kepler's laws of planetary motion EXCEPT?
a.) Planets follow elliptical orbits with the sun at one of its foci
b.) The period of a planet (time it takes to go around the sun) is related to its distance from its Sun
c.) The period of a planet(time it takes to go around the sun) is related to the planet's mass
Answers
C. The period of a planet (time it takes to go around the sun) is not related to its mass.
The first law states that the planets follow elliptical orbits with the sun at one of its foci, the second law states that an imaginary line drawn from the sun to a planet sweeps out equal areas in equal times, and the third law states that the square of the period of a planet is directly proportional to the cube of its average distance from the sun. Since the period of a planet is not related to its mass, answer C is not one of Kepler's laws of planetary motion.
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A motorcycle travels, in one direction only, with an average speed of 916.66 m/min during the first 30 minutes of its travel and 900 m/min during the next 20 minutes.
Calculate (Units must be in units of the international system):
a. The total distance traveled.
b. The average speed.
Answers
Explanation:
a. To calculate the total distance traveled, we need to find the distance traveled during the first 30 minutes and the distance traveled during the next 20 minutes, and then add them up.
During the first 30 minutes:
distance = speed × time
distance = 916.66 m/min × 30 min
distance = 27,499.8 meters
During the next 20 minutes:
distance = speed × time
distance = 900 m/min × 20 min
distance = 18,000 meters
Total distance traveled:
distance = 27,499.8 meters + 18,000 meters
distance = 45,499.8 meters
Therefore, the total distance traveled is 45,499.8 meters.
b. To calculate the average speed, we need to divide the total distance traveled by the total time taken.
Total time taken:
time = 30 min + 20 min
time = 50 min
Average speed:
speed = distance ÷ time
speed = 45,499.8 meters ÷ 50 min
speed = 909.996 m/min
Therefore, the average speed is 909.996 m/min.
A 15.0 cm tall object is placed 35.0 cm from a convex lens, which
has a focal length of 15.0 cm. Calculate the height for the image.
Answers
The image is reversed if the image height has a negative sign. The image is inverted and has a height of -6.45 cm.
How is the image's height determined?The lens equation describes how a convex lens's object distance (d o), image distance (d i), and focal length (f) relate to one another:
1/f = 1/d_o + 1/d_i
1/d_i = 1/15.0 cm - 1/35.0 cm 1/d_i = 1/f - 1/d o
1/d_i = 0.0667 cm -1
d_i = 15.0 cm
The image's magnification (M) is determined by:
M = - d_i / d_o
M = -15.0 cm / 35.0 cm
M = -0.43
M = h_i / h_o
h_i = M x h_o
The value for h_o is 15.0 cm. With M = -0.43, we obtain:
h_i = -0.43 * 15.0 cm
h_i = -6.45 cm
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A 1450 car having a speed of 25.2 collides with a 7.5 truck moving in the same direction at 20.0 . Velocity of the car after the collision changed to 15.0 in the initial direction. What is the velocity of the truck after the collision?
Answers
The velocity of the truck after the collision, given that a 1450 kg car having a speed of 25.2 m/s collides with the 7.5 kg truck is 1992 m/s
How do I determine the velocity of the truck after collision?The following data were obtained from the question given above:
Mass of car (m₁) = 1450 KgInitial velocity of car (u₁) = 25.2 m/sMass of truck (m₂) = 7.5 KgInitial velocity of truck (u₂) = 20.0 m/sFinal velocity of car (v₁) = 15.0 m/sFinal velocity of truck (v₂) = ?The velocity of the truck after the collision can be obtained by using the law of conservation of linear momentum as follow:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(1450 × 25.2) + (7.5 × 20) = (1450 × 15) + (7.5 × v₂)
36540 + 150 = 21750 + (7.5 × v₂)
36690 = 21750 + 7.5v₂
Collect like terms
7.5v₂ = 36690 - 21750
7.5v₂ = 14940
Divide both sides by 7.5
v₂ = 14940 / 7.5
v₂ = 1992 m/s
Thus, the velocity of the truck after collision is 1992 m/s
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A sled with a mass 47.0 kg is pulled along flat, snow-covered ground. The static friction coefficient is 0.30, and the kinetic friction coefficient is 0.10.
(a) What does the sled weigh?
(b) What force will be needed to start the sled moving?
(c) What force is needed to keep the sled moving at a constant velocity?
(d) Once moving, what total force must be applied to the sled to accelerate it at 3.3 m/s2?
Answers
(a) The weight of the sled is given by the product of its mass and the acceleration due to gravity:
w = m*g = 47.0 kg * 9.81 m/s^2 = 461.07 N
(b) The force needed to start the sled moving is the maximum force of static friction, which is given by:
f_s = μ_s * w = 0.30 * 461.07 N = 138.32 N
(c) Once the sled is moving at a constant velocity, the force needed to keep it moving is equal to the force of kinetic friction, which is given by:
f_k = μ_k * w = 0.10 * 461.07 N = 46.11 N
(d) The total force needed to accelerate the sled at 3.3 m/s^2 is the sum of the force of kinetic friction and the force required to produce the desired acceleration:
F = f_k + m*a = 46.11 N + 47.0 kg * 3.3 m/s^2 = 200.97 N
A ray of light in water has a wavelength of 4.42×
[tex] {10}^{ - 7} m[/tex]
What is the wavelength of that way while passing through ice?
Answers
When light is travelling through ice, its wavelength is 4.50 10⁻⁷ metres.
When light travels from air to glass, what happens to its wavelength?Since glass has a higher index of refraction than air, light travels more slowly through glass than through air (n=c/v). Although the wavelength does not change, the frequency does because the speed does.
The following equation describes the relationship between the wavelengths of light in two different media:
n1 * λ1 = n2 * λ2
We may rewrite this equation to get the wavelength of the light in the second medium while assuming that the light's frequency stays constant: λ2 = (n1 / n2) * λ1
For water, the refractive index is about 1.333, and for ice, it is about 1.31. Therefore, we have:
λ2 = (1.333 / 1.31) * 4.42×10⁻⁷ m
λ2 = 4.50×10⁻⁷ m
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An ideal gas undergoes an isothermal (constant-temperature) compression at temperature T, with its volume changing from V1 to V2. (i) Prove that the work done during this process is given by an expression =12. (5) (ii) Is the work done positive or negative? Give reasons.
Answers
The effort required to compress an ideal gas is therefore given by W = -1/2 nRT ln(V2/V1) during an isothermal process.
What labour is involved in an ideal gas's isothermal expansion and compression?When an ideal gas is subjected to isothermal expansion (T = 0) in a vacuum, the work done is equal to zero as pex=zero. Joule established q = 0 empirically; hence, U = 0. Equation 1 can be written as: for both reversible and irreversible isothermal changes. Reversible isothermal change q = -w = pex (Vf-Vi)
nRT = ln(V2/V1) - W
If we solve for W, we obtain:
W = ln(V1/V2)nRT (v) If we condense this phrase, we get:
W=-nRT ln(V2/V1).
W=-RT ln(V2/V1).
W=-1/2nRT ln(V2/V1)
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Hanif is the tallest player in a volleyball team. He is in spiking position when Johan delivers him the ball. The 0.226-kg volleyball is 2.29 m above the ground and has a speed of 1.06 m/s. Hanif spikes the ball, doing 9.89 J of work on it. (a) Determine the potential energy and the kinetic energy of the ball before Hanif spikes it. (b) The total mechanical energy of the ball before Hanif spikes it.
Answers
(a)
Potential Energy = 4.95 J
Kinetic Energy =0.132 J
b.) the total mechanical energy of the ball before Hanif spikes it is 5.08 J.
How to calculate?Potential energy (PE) = mgh
Kinetic energy (KE) = (1/2)mv^2
m = 0.226 kg
h = 2.29 m
v = 1.06 m/s
g = 9.81 m/s^2 (acceleration due to gravity)
PE = mgh = (0.226 kg)(9.81 m/s^2)(2.29 m) = 4.95 J
KE = (1/2)mv^2 = (1/2)(0.226 kg)(1.06 m/s)^2 = 0.132 J
(b) The total mechanical energy of the ball before Hanif spikes it is the sum of its potential and kinetic energy:
Total mechanical energy = PE + KE = 4.95 J + 0.132 J = 5.08 J
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Which of the following is an example of an object with only gravitational potential energy?
Group of answer choices
A book resting on a shelf.
A ball thrown straight upwards
A bone lying on the floor
A fruit falling down off of a tree.
Answers
Since they are not at a height above the ground where gravity can act on them, A ball thrown straight upwards and a bone on the floor both have zero gravitational potential energy.
Of the objects, which one contains gravitational potential energy?If an object is placed at a height above (or below) the zero height, it has gravitational potential energy. If an object is not in its equilibrium position on an elastic material, it has elastic potential energy.
Gravitational potential: What is it?The term gravitational potential energy refers to the energy that an item stores as a result of its elevation above the Earth's surface. This energy is a result of an object being subjected to gravity. EP=mgh.
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PLEASE HELP ME! LIKE ASAP! Imagine a population of bugs that has two traits for body color. Some bugs are bright. Some are dark. A new predator can see the bright bugs more easily than the dark bugs. Describe how natural selection could affect this trait in the bug population over time.
Answers
Answer: if the predator sees the light bugs easier than the dark bugs then the bright bug will most likely go extinct
Explanation:
Uses of transistor?
Answers
Answer:
Transistors and Their Uses
Transistors are electronic devices that are used to amplify and switch electronic signals. They are widely used in various applications, from small electronic devices to large industrial systems. Here are some common uses of transistors:
1. Amplification: Transistors are commonly used in audio amplifiers, where they amplify weak audio signals to produce a louder sound. They are also used in radio and television receivers to amplify the weak signals received from antennas.
2. Switching: Transistors are used as switches in electronic circuits, where they can be turned on or off to control the flow of current. They are commonly used in digital circuits, where they can be used to turn on or off individual bits of data.
3. Voltage Regulation: Transistors can be used as voltage regulators, where they can be used to regulate the output voltage of a power supply. They are commonly used in electronic devices such as computers and televisions, where a stable voltage supply is required.
4. Oscillation: Transistors can be used in oscillator circuits to produce a steady periodic waveform, such as a sine wave. These circuits are commonly used in electronic devices such as radios and televisions.
5. Logic Gates: Transistors are used in logic gates, which are the building blocks of digital circuits. They can be used to implement Boolean logic functions such as AND, OR, and NOT.
6. Memory: Transistors are used in memory circuits, such as dynamic random-access memory (DRAM), where they are used to store data. DRAM is commonly used in computers as the main memory.
7. Power Control: Transistors can be used in power control circuits, where they can be used to control the amount of power delivered to a load. They are commonly used in electronic devices such as motor controllers and power supplies.
In conclusion, transistors are versatile devices that are used in a wide range of electronic applications. They can be used for amplification, switching, voltage regulation, oscillation, logic gates, memory, and power control. Transistors have revolutionized the field of electronics and have enabled the development of many modern electronic devices.
Answer:
they used to transition of current and flow of them like amplifier
A double-convex thin lens has surfaces with equal radii of curvature of magnitude 3.00 cm. Looking through this lens, you observe that it forms an image of a very distant tree at a distance of 1.98 cm from the lens. What is the index of refraction of the lens?
Answers
n = 1.50. In this case, the index of refraction is 1.50, which demonstrates that the lens is able to refract light in order to form an image.
Given ParametersRadii of curvature (r1, r2): 3.00 cm Distance of image (d): 1.98 cmLet us calculate the focal length of the lens:
F = 1/[(1/r1)+(1/r2)]
F = 1/[(1/3.00 cm)+(1/3.00 cm)]
F = 1.50 cm
Calculate the index of refraction:
n = 1/(1/f - 1/d)
n = 1/(1/1.50 cm - 1/1.98 cm)
n = 1.50
Both of these surfaces cause light rays to bend and converge at a focal point on the other side of the lens. The index of refraction of the lens can be calculated by using the equation n = 1/(1/f - 1/d), where f is the focal length and d is the distance of the image.
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We perform the experiment with the same rod employed in the previous experiment, but now the rod is mounted on our smart cart. The cart is pushed with maximum acceleration a. Justify based on theory the maximum displacement angle observed in both runs. Watch it here and get the data here. [Hint: There is common ground between this question and section 3 in lab 2].
Answers
The maximum displacement angle observed in the experiment can be calculated based on the acceleration of the cart and the value of g.
What is Displacement Angle?
Displacement angle is the angle through which an object has moved or rotated with respect to a reference point or position. In the case of a pendulum, it refers to the maximum angle the pendulum swings away from its vertical position before reversing direction due to the force of gravity.
The maximum displacement angle observed in the experiment depends on the initial velocity of the pendulum and the acceleration of the cart. When the cart is pushed with maximum acceleration a, the pendulum initially experiences a force due to its inertia, which causes it to move at an angle. The angle of displacement is directly proportional to the initial velocity of the pendulum.
Based on the theory of pendulums, the maximum angle of displacement is given by:
θ = arcsin(a/g)
Where θ is the maximum angle of displacement, a is the acceleration of the cart, and g is the acceleration due to gravity (approximately 9.81 m/[tex]s^{2}[/tex]).
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