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3 years ago

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A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apar

t. For the force that each object exerts on the other to be a maximum, what should q be? a) q 20 c) q Q/2 d) q /4

1 answer:

Whitepunk [10]3 years ago

4 0

Answer:

q = Q/2

option c is correct

Explanation:

given data

charge = Q

distance r = 1 m

charge = q

to find out

what should q be

solution

we have given charge q is remove so

q1 = Q-q

q2 = q

we know by coulombs law force

force = kq1q2 / r² .............1

put here value and we know k is constant

F = k (Q-q) q / 1

now take derivative charge q and put = 0

dF/dq = k d(Q-q)q / dq

so k (Q-q)q =0

Q = 2q

q = Q/2

so option c is correct

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To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: $V_{f}=V_{i}+gt$
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$V_{f}$ is the final velocity
$V_{i}$ is the initial velocity
$g$ is the acceleration due to gravity $32 \frac{ft}{s^2}$
$t$ is the time
2. Kinematic equation for distance: $d=V_{i}t+ \frac{1}{2} gt^2$
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$V_{i}$ is the initial velocity
$V_{f}$ is the final velocity
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$t$ is the time

First, we are going to convert 21.82 mi/h to ft/s:
$21.82 \frac{mi}{h} =31.21 \frac{ft}{s}$

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: $V_{f}=0$ . We also know that it initial speed is 31.21 ft/s, so $V_{i}=31.21$ . Lets replace those values in our formula to find $t$ :
$V_{f}=V_{i}+gt$
$0=31.21+(-32)t$
$-32t=-31.21$
$t= \frac{-31.21}{-32}$
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Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
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$d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2$
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Now we can add the height of the building and the maximum height of the object:
$d=160+15.22=175.22$ ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
$d=V_{i}t+ \frac{1}{2} gt^2$
$175.22=31.21t+ \frac{1}{2} (32)t^2$
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$V_{f}=31.21+(32)(2.47)$
$V_{f}=110.25$ ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s

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