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kow [346]
3 years ago
12

In a thundercloud there may be an electric charge of +40 C near the top of the cloud and -40 C near the bottom of the cloud. The

se charges are separated by about 2.0 km. What is the electric force between these two sets of charges? (ke = 8.99 ´ 109 N·m2/C2)
Physics
1 answer:
Ratling [72]3 years ago
5 0

Answer:

Electric force, F=-3.59\times 10^6\ N

Explanation:

Given that,

Electric charge 1, q_1=+40\ C

Electric charge 2, q_2=-40\ C

Distance, d=2\ km=2\times 10^3\ m

To find,

The electric force between these two sets of charges.

Solution,

There exists a force between two electric charges and this force is called electrostatic force. It is equal to the product of electric charged divided by square of distance between them.

F=k\dfrac{q_1q_2}{d^2}

k is the electrostatic constant

F=8.99\times 10^9\times \dfrac{40\times (-40)}{(2\times 10^3)^2}

F=-3.59\times 10^6\ N

So, the electric force between these two sets of charges is -3.59\times 10^6\ N.

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A bottle lying on the windowsill falls off and takes 4.95 seconds to reach the ground. The distance from the windowsill to the g
Liula [17]
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           Distance  =  (1/2 acceleration of gravity) x (square of the falling time)

We want to see how the time will be affected 
if  ' D ' doesn't change but ' g ' does. 
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Divide each side by ' g ' :      2 D/g =                  t² 

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They don't change by the same factor, because  1/g  is inside the square root.  So 't' changes the same amount as  √1/g  does.

Gravity on the surface of the moon is roughly  1/6  the value of gravity on the surface of the Earth.

So we expect ' t ' to increase by  √6  =  2.45 times.

It would take the same bottle  (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
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