Please use Gauss’s law to find the electric field strength E at a distance r from the center of a sphereof radius R with volume charge density ???? = cr 3 and total charge ????. Your answer should NOT contain c. Be sure to consider regions inside and outside the sphere.

Answers

Answer 1

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

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Does lighting striking the earth considered the speed of light?
A triply charged ion with velocity 7.00 × 10^6 m/s moves in a path of radius 36.0 cm in a magnetic field of 0.55 T in a mass spectrometer. What is the mass of this ion?
A farmer is using a rope and pulley to lift a bucket of water from the bottom of a well. the farmer uses a force f1=57.5 n to pull the bucket of water upwards. the total mass of the bucket of water is f2= 3.9kg. -Calculate how much work Wg in J gravity does on the bucket filled with water as the farmer lifts it up the well. -Calculate the net work Wnet in J done on the bucket of water by the two forces F1 and Fg.
Let A be the last two digits, and let B be the last three digits, and the C be the sum of the last 4 digits of your 8-digit student ID. (Example: For 20245347, A = 47, B = 347, and C = 19) A train moves at an average speed of (23.0 + A) m/s for (250.0 + B) seconds and then at an average speed of (45.0 + C) m/s for (800.0 + B) seconds. Determine the average speed for the entire time in meters per second (m/s). Round your final answer to 3 significant figures.

For exercise, an athlete lifts a barbell that weighs 400 N from the ground to a height of 2.0 m in a time of 1.6 s. Assume the efficiency of the human body is 25%, and that he lifts the barbell at a constant speed. Show all work and include proper unit for your final answer.a) In applying the energy equation (ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W) to the system consisting of the earth, the barbell, and the athlete,
1. Which terms (if any) are positive?
2. Which terms (if any) are negative?
3. Which terms (if any) are zero?
b) Determine the energy output by the athlete in SI unit.
c) Determine his metabolic power in SI unit.
d) Another day he performs the same task in 1.2 s.
1. Is the metabolic energy that he expends more, less, or the same?
2. Is his metabolic power more, less, or the same?

Answers

Answer:

Explanation:

(ΔK + ΔUg + ΔUs + ΔEch + ΔEth = W)

ΔK is increase in kinetic energy . As the athelete is lifting the barbell at constant speed change in kinetic energy is zero .

ΔK = 0

ΔUg is change in potential energy . It will be positive as weight is being lifted so its potential energy is increasing .

ΔUg = positive

ΔUs is change in the potential energy of sportsperson . It is zero since there is no change in the height of athlete .

ΔUs = 0

ΔEth is change in the energy of earth . Here earth is doing negative work . It is so because it is exerting force downwards and displacement is upwards . Hence it is doing negative work . Hence

ΔEth = negative .

b )

work done by athlete

= 400 x 2 = 800 J

energy output = 800 J

c )

It is 25% of metabolic energy output of his body

so metalic energy output of body

= 4x 800 J .

3200 J

power = energy output / time

= 3200 / 1.6

= 2000 W .

d )

1 ) Since he is doing same amount of work , his metabolic energy output is same as that in earlier case .

2 ) Since he is doing the same exercise in less time so his power is increased . Hence in the second day his power is more .

Final answer:

Positive, negative, and zero terms in the energy equation. Calculation of energy output and metabolic power. Comparison of metabolic energy and power for different time durations.

Explanation:

To apply the energy equation to the system, we need to determine whether each term is positive, negative, or zero:

Positive terms:

ΔUg - the change in gravitational potential energy is positive as the barbell is lifted vertically from the ground.
ΔUs - the change in elastic potential energy is positive if there is any stretch or compression in the system.

Negative terms:

ΔK - the change in kinetic energy is negative as the barbell is lifted at a constant speed, so there is no change in velocity.
ΔEch - the change in chemical potential energy is negative if the athlete is not ingesting any food or drinks during the exercise.

Zero terms:

ΔEth - the change in thermal energy is zero if there is no heat transfer in the system.

To determine the energy output by the athlete, we can calculate the work done on the barbell using the formula W = ΔUg. In this case, the work done is equal to the change in gravitational potential energy, which is equal to mgh. Thus, W = 400 N × 2.0 m = 800 J. So the energy output by the athlete is 800 J.

The metabolic power can be calculated using the equation P = W / t, where P is the power, W is the work done, and t is the time taken. Substituting the given values, P = 800 J / 1.6 s = 500 W. Therefore, the metabolic power of the athlete is 500 W. If the task is performed in a faster time, the metabolic energy expended will be the same. However, the metabolic power will be greater as the work is done in less time.

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A 50 kg child on a skateboard experiences a 75-N force as shown.What is the expected acceleration of the child?

F = 75 N
7
.

A. 0.67 m/s2
B. 1.50 m/s2
C. 6.70 m/s2
D. 25.0 m/s2

Answers

Answer:

1.50 m/s²

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = (f)/(m) \n$

f is the force

m is the mass

From the question we have

$a = (75)/(50) = (3)/(2) \n$

We have the final answer as

1.50 m/s²

Hope this helps you

Assume that the speed of light in a vacuum has the hypothetical value of 18.0 m/s. A car is moving at a constant speed of 14.0 m/s along a straight road. A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s. How much time does the driver of the car measure for his trip between the poles?

Answers

Answer:

Observed time, t = 5.58 s

Explanation:

Given that,

Speed of light in a vacuum has the hypothetical value of, c = 18 m/s

Speed of car, v = 14 m/s along a straight road.

A home owner sitting on his porch sees the car pass between two telephone poles in 8.89 s.

We need to find the time the driver of the car measure for his trip between the poles. The relation between real and observed time is given by :

$T=\frac{t}{\sqrt{1-(v^2)/(c^2)} }$

t is observed time.

$t=T* \sqrt{1-(v^2)/(c^2)} \n\nt=8.89* \sqrt{1-(14^2)/(18^2)} \n\nt=5.58\ s$

So, the time observed by the driver of the car measure for his trip between the poles is 5.58 seconds.

Ken Griffey, Jr's warehouse shot in the 1933 home run derby travelled 93 feet per second for 5 seconds. How far did he hit the ball?

Answers

Answer:

465 feet because 93*5 = 465, btw that was 1993 not 1933

Explanation:

Dave rows a boat across a river at 4.0 m/s. the river flows at 6.0 m/s and is 360 m across.a. in what direction, relative to the shore, does dave’s boat go?
b. how long does it take dave to cross the river?
c. how far downstream is dave’s landing point?
d. how long would it take dave to cross the river if there were no current?

Answers

a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.

Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

$\theta= arctan((v_y)/(v_x))=arctan((4.0 m/s)/(6.0 m/s))=arctan(0.67)=33.7^(\circ)$

relative to the direction of the river.

b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

$t=(S_y)/(v_y)=(360 m)/(4.0 m/s)=90 s$

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

$S_x = v_x t =(6.0 m/s)(90 s)=540 m$

so, Dave's landing point is 540 m downstream.

d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).

Crude oil is a mixture of many different components. The extraction of crude oil from the Earth is important, but its refinement into different substances is a key piece to obtaining as many uses as possible from the crude oil. Using the diagram, justify the source of data used to develop the technology for refining the crude oil.

Answers

Crude oil is a mixture of nitrogen, oxygen, sulphur, and hydrogen components

What is Crude oil?

A combination of hydrocarbons known as crude oil is one that is found in naturally occurring subsurface reservoirs in the liquid phase and continues to be liquid at atmospheric pressure after passing through surface separation equipment.

Refineries transform crude oil into useful products including gasoline, diesel, and aviation fuels for transportation. Gasoline: A fuel used in both personal and commercial vehicles that are made for internal combustion engines.

In addition to some nitrogen, sulphur, and oxygen, crude oil is a combination of very flammable liquid hydrocarbons (compounds mostly made of hydrogen and carbon).

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Answer: Different fuel components boil at different temperatures, allowing them to be separated.

Explanation:

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