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3 years ago

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Two identical point charges are fixed to diagonally opposite corners of a square that is 0.715 m on a side. Each charge is +4.0

x 10-6 C. How much work is done by the electric force as one of the charges moves to an empty corner?

1 answer:

siniylev [52]3 years ago

5 0

Answer:

Work done by the electric force as one of the charges moves to an empty corner is 25.71 KJ

Explanation:

Work done on a unit positive charge = Electric field potential X distance moved by the charge.

Electric field potential = (Kq)/r²

Where;

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is magnitude of point charge = 4.0 x 10⁻⁶ C

r is the distance between the point charges = 0.715 m

Then, Work done on a unit positive charge = (Kq)/r² X r

Work done on a unit positive charge = (kq)/r

Work done on a unit positive charge = (8.99 X 10⁹ X 4.0 x 10⁻⁶)/0.715

Work done on a unit positive charge = 25.71 X 10³ J = 25.71 KJ

Therefore, work done by the electric force as one of the charges moves to an empty corner is 25.71 KJ

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Given :

Initial velocity , u = 0 m/s² .

To Find :

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Since, acceleration is constant.

Using equation of motion :

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3 years ago

A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w

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Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

$\displaystyle P=\frac{W}{t}$

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

$W=F.x$

The second Newton's law gives us the net force as

$F=m.a$

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

$v_f=v_o+at$

$v_f=at$

Solving for a

$\displaystyle a=\frac{v_f}{t}={5.4}{1}$

$a=5.4\ m/s^2$

The distance traveled in the interval is given by

$\displaystyle x=v_o.t+\frac{a.t^2}{2}$

Since vo=0

$\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}$

$x=2.7\ m$

The force is given by

$F=m.a$

We don't know the value of m, so the force is

$F=2.7m$

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$W=F.x=2.7m(5.4)$

$W=14.58m$

The power is finally computed

$\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}$

$P=14.58m$

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

$\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}$

$a=8.8\ m/s^2$

The distance is

$\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}$

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The net force is

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The work done by the sprinter is now computed as

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At last, the output power is

$\displaystyle P=\frac{33.44m}{0.5}=66.88m$

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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