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ss7ja [257]
3 years ago
11

A stone is thrown horizontally with an initial speed of 10 m/s from the edge of a cliff. A stopwatch measures the stone's trajec

tory time from the top of the cliff to the bottom to be 4.3 s. What is the height of the cliff if air resistance is negligibly small?
Physics
1 answer:
olga2289 [7]3 years ago
5 0

Answer:

The height of the cliff is 90.60 meters.

Explanation:

It is given that,

Initial horizontal speed of the stone, u = 10 m/s

Initial vertical speed of the stone, u' = 0 (as there is no motion in vertical direction)

The time taken by the stone from the top of the cliff to the bottom to be 4.3 s, t = 4.3 s

Let h is the height of the cliff. Using the second equation of motion in vertical direction to find it. It is given by :

h=u't+\dfrac{1}{2}gt^2

h=\dfrac{1}{2}gt^2

h=\dfrac{1}{2}\times 9.8\times (4.3)^2

h = 90.60 meters

So, the height of the cliff is 90.60 meters. Hence, this is the required solution.

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What happens at night- describing air circulation
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5 0
3 years ago
2. An elephant puts a force of 30,000 N on its four feet, which have a
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8 0
3 years ago
Two spheres have identical charges and are 75 cm apart. the force between them is +0.30 n. what is the magnitude of the charge o
ella [17]

Answer:

4.33\cdot 10^{-6}C, charges are both positive or both negative

Explanation:

The electrostatic force between the two spheres is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1 and q2 are the charges on the two spheres

r is the distance between the centres of the two spheres

In this problem, we have

F=+0.30 N is the force

r=75 cm=0.75 m is the distance between the spheres

q_1 =q_2 =q because the two spheres have identical charge

Solving the formula for q, we find

q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(+0.30 N)(0.75 m)^2}{9\cdot 10^9}}=4.33\cdot 10^{-6}C

And the two charges have the same sign (so, both positive or both negative), since the sign of the force is positive (+0.30 N), so it is a repulsive force.

5 0
3 years ago
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