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3 years ago

How should the distance, D, between two point charges, q1 and q2, be changed to double their electric potential energy

1 answer:

6 0

The electrostatic potential energy, U, of one point charge q at position d in the presence of an electric field E is defined as the negative of the work W done by the electrostatic force to bring it from the reference position d to that position

$u \: = \: \frac{kq1q2}{d}$

Thus, to double the electric potential energy U we need to reduce the distance of separation by half (1/2) because they are inversely proportion

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At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is

Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

$\dfrac{1}{2}mv^2=mgl$

$v=\sqrt{2gl}$

Put the value into the formula

$u=\sqrt{2\times9.8\times50.0\times10^{-2}}$

$u=3.13\ m/s$

The initial speed of the ball $u_{1}=3.13\ m/s$

The initial speed of the block $u_{2}=0$

(a). We need to calculate the speed of the ball after collision

Using formula of collision

$v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13$

$v_{1}=-2.01\ m/s$

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

$v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}$

Put the value into the formula

$v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0$

$v_{2}=1.11\ m/s$

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0

3 years ago

A stationary charge is located between the poles of a horseshoe magnet. Is a magnetic force excerted of the charge? why?

vredina [299]

A stationary charge is located between the poles of a horseshoe magnet. The magnetic force exerted by the charge is zero.

<h3>What is charge?</h3>

Charge is the physical property of matter which cause a particle to attract or repel when placed in its field.

A stationary charged particle does not interact with a static magnetic field. A charge placed in a magnetic field experiences a magnetic force. There will be no magnetic force acting on a stationary charge. The charge must be moving in order to have magnetic force on it.

Thus, the magnetic force exerted by the charge is zero.

Learn more about charge.

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1 year ago

You are trying to find the amount of heat transferred between two substances. In order to do this, you plan

Zarrin [17]

Answer:

do you have kids ?

Explanation:

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2 years ago

What is one disadvantage of a series circuit?

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If one bulb goes out then all the others won't light up because electricity will be cut off. It's a disadvantage because in a parallel circuit if one bulb burns out all the others will still be on because they won't be affected. I hope I've helped you ☺

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3 years ago