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yan [13]
3 years ago
15

Greg dropped a coin from the h meter tall building. If it takes t seconds to reach the ground, determine the position of the coi

n at the t/2 seconds
Physics
1 answer:
My name is Ann [436]3 years ago
8 0

consider the motion of the coin in vertical direction :

Assuming down direction as positive and top of building as origin

Y₀ = initial position of displacement of the coin at the top of building = 0 m

Y = final position of displacement of the coin as it hits the ground = h

t = time taken to hit the ground

a = acceleration = acceleration due to gravity = g

v₀ = initial velocity at the top of the building = 0 m/s

using the equation

Y = Y₀ +  v₀ t + (0.5) a t²

h = 0 + 0 t + (0.5) g t²

h = (0.5) g t²                                                 eq-1


consider the motion of the coin for time "t/2"

Y₀ = initial position of displacement of the coin at the top of building = 0 m

Y' = final position of the coin after time "t/2"

t' = time of travel = t/2

a = acceleration = acceleration due to gravity = g

v₀ = initial velocity at the top of the building = 0 m/s

using the equation

Y' = Y₀ +  v₀ t' + (0.5) a t'²

Y' = 0 + 0 t + (0.5) g (t/2)²

Y' = (0.5) gt²/4

using eq-1

Y' = h/4                           since h = (0.5) g t²


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Answer:

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A = \pi r^2 = \frac{\pi d^2}{4} = \frac{\pi \times (0.1)^2}{4} = 0.00786 \ m^2

L = \frac{(1000)^2 \times (4\pi \times 10^{-7}) \times (0.00786)}{0.45} \\\\L = 0.02195 \ H\\\\L = 21.95 \ mH

(b) The energy stored in the inductor when 21 A current ;

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Explanation:

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