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Genrish500 [490]
2 years ago
13

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

ctric field of N/C at point P. Determine the work required of an external agent to move a 3.0-μC charge along the x axis from infinity to point P without any net change in the kinetic energy of the particle.
Physics
1 answer:
Julli [10]2 years ago
4 0

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C is the magnitude of the charge

\Delta V = 6.0 kV = 6000 V is the potential difference between point P and infinity

Substituting into the equation, we find

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

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Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart
Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0
2 years ago
Consider the position vs. time graph below for a woman's movement in a hallway. What is the woman's velocity from 4 to 5 s?
Ksenya-84 [330]

Answer:

The answer is "6\  \frac{m}{s}"

Explanation:

The formula for velocity:

\to \overline{v}={\frac{\Delta x}{\Delta t}}

      =\frac{6}{1}\\\\=6\  \frac{m}{s}

7 0
1 year ago
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
madreJ [45]
In order to answer this exercise you need to use the formulas

 S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s
7 0
2 years ago
Read 2 more answers
A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m
Inessa [10]

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

Mass of ball = 0.65 kg

Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

(0.65 + 54)Vf = 54.65Vf

Where Vf is their final common velocity.

Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

4 0
1 year ago
What is the frequency of radiation whose wavelength is 11.5 a0 ?
irakobra [83]

Answer:

The frequency of radiation is 2.61 \times 10^{17} s^{-1}

Explanation:

Given:

Wavelength \lambda = 11.5 \times 10^{-10} m

Speed of light c = 3 \times 10^{8} \frac{m}{s}

For finding the frequency of radiation,

  c = f \lambda

  f = \frac{c}{\lambda}

  f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }

  f = 2.61 \times 10^{17} s^{-1}

Therefore, the frequency of radiation is 2.61 \times 10^{17} s^{-1}

4 0
2 years ago
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