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4 years ago

a bat experts a 120 N force on a 0.134 kg baseball for 0.04 seconds. what impolse is excerted on the bat

1 answer:

3 0

The impulse exerted by a force F on an object in a time $\Delta t$ is given by
$I = F \Delta t$
In our problem, we have a force of 120 N: F=120 N, applied to the baseball for a time of $\Delta t=0.04 s$ . Therefore the impulse is
$I=F \Delta t = (120 N)(0.04 s)=4.8 Ns$

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A parallel-plate capacitor in air has circular plates of radius 2.8 cm separated by 1.1 mm. Charge is flowing onto the upper pla

Hatshy [7]

Answer:

The time rate of change of the electric field between the plates is $\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$

Explanation:

From the question we are told that

The radius is $r = 2.8 \ cm = 0.028 \ m$

The distance of separation is $d = 1.1 \ mm = 0.0011 \ m$

The current is $I = 5 \ A$

Generally the electric field generated is mathematically represented as

$E = \frac{q }{ \pi * r^2 \epsilon_o }$

Where $\epsilon_o$ is the permitivity of free space with a value

$\epsilon_o = 8.85*10^{-12 }\ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

So the time rate of change of the electric field between the plates is mathematically represented as

$\frac{E }{t} = \frac{q}{t} * \frac{1 }{ \pi * r^2 \epsilon_o }$

But $\frac{q}{t } = I$

$\frac{E }{t} = * \frac{I }{ \pi * r^2 \epsilon_o }$

substituting values

$\frac{E }{t} = * \frac{5 }{3.142 * (0.028)^2 * 8.85 *10^{-12} }$

$\frac{E }{t} = 2.29 *10^{14} \ N \cdot C \cdot s^{-1}$

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4 years ago

Might it be possible to explain the interaction of the rod and pieces of paper as a gravitational interaction? please explain wh

mina [271]

It is not possible to explain the interaction of the rod and pieces of paper as a gravitational interaction.

<h3>What is Gravitational interaction?</h3>

This is defined as the interaction between a particle or body resulting from their mass. This type of interaction is usually weak and occurs in all distances possible.

It is not gravitational interaction, because the rod attracts paper only against the gravitational force of the earth and there is no attraction between both bodies under a different condition.

This is therefore the reason why it is not possible to explain the interaction of the rod and pieces of paper as a gravitational interaction.

Read more about Gravitational interaction here brainly.com/question/25624188

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2 years ago

Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e

NeTakaya

We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

$Q=\frac{E}{Nr^2}$

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the <em>magnitude </em>of charge moved, Q is mathematically given as

$Q=\frac{E}{Nr^2}$

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

$Q=\frac{E}{Nr^2}$

For more information on this visit

brainly.com/question/16517842

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