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vagabundo [1.1K]
3 years ago
5

Is it true that mproving overall fitness may not be possible for people who have physical limitations.

Physics
1 answer:
OleMash [197]3 years ago
8 0
That's not true people that have a broken leg use crutches and the constant lifting of their own body weight will help them gain muscle 
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A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/
hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is 9.1\cdot 10^{-4} J

C) The work done by the electric field is -2.4\cdot 10^{-3} J

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the particle

\theta is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

  • The force is directed vertically upward (because the field is directed vertically upward)
  • The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

\theta=90^{\circ}

and cos 90^{\circ}=0: therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

\theta=0^{\circ}

and

cos 0^{\circ}=1

Therefore, the work done by the electric force is

W=Fd

and we have:

F=qE is the magnitude of the electric force. Since

E=4.30\cdot 10^4 V/m is the magnitude of the electric field

q=32.0 nC = 32.0\cdot 10^{-9}C is the charge

The electric force is

F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

\theta=90^{\circ}+45^{\circ}=135^{\circ}

Moreover, we have:

F=1.38\cdot 10^{-3} N (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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What is the difference between each distance traveled and displacement travled
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Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.

While, Distance is a scalar magnitude that measures over the trajectory.

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A small, positively charged ball is moved close to a large, positively charged ball. Which describes how the small ball likely r
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Answer:c

Explanation:

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A child standing on a bus remains still when the bus is at rest. When the bus moves forward and then slows down, the child conti
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Newton's first law of motion
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Help a girl out ?! Plz ill give brainlest
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Answer:

I am pretty sure it is B.

Explanation:

I hope this helped if it didn't I am truly sorry

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