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Orlov [11]
3 years ago
5

A stone is dropped from a height of 49m and simultaneously another ball is thrown upward from the ground with a speed of 40m/s.

when and where do the two stone meet
​
Physics
1 answer:
Lilit [14]3 years ago
5 0

Answer:

S1 = 1/2 g t^2     distance stone falls in time t

S2 = Vy t - 1/2 g t^2   distance thrown stone rises in time t

H = 49 = S1 + S2 = Vy t

t = 49 / 40 sec   time when stones meet

Check:

Stone 1 falls:      1/2 g t^2 = 1/2 * 9.8 * (49 / 40)^2 = 7.35 m

Stone 2 rises :  40 * (49 / 40) - 1/2 * 9.8 (49 / 40)^2 = 41.65 m

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Answer:

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(b) The energy stored, E = 4.84 J

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Explanation:

(a) Self inductance is calculated as;

L = \frac{N^2 \mu_0 A}{l}

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

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

E = \frac{1}{2}LI^2\\\\E = \frac{1}{2} \times (0.02195) \times (21) ^2\\\\E = 4.84 \ J

(c) time it can be turned off if the induced emf cannot exceed 3.0 V;

emf = L \frac{\Delta I}{\Delta t} \\\\t = \frac{LI}{emf} \\\\t = \frac{0.02195 \times 21}{3} \\\\t = 0.154 \ s

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