Question

water drops from a faucet at the rate of 5 drops/sec. what is the position of the lower drop if it has attained a velocity of 3m/s? what is the position of the upper drop if the lower drop has attained a velocity of 3m/s? what is the distance between two consecutive drops after the lower drop attains a velocity of 3m/s?

Answer 1

Answer:

distance between two drops is 43.75 cm

Explanation:

Velocity of the lower drop is given as

v = 3 m/sv=3m/s

so the time taken to reach this speed

v = gtv=gt

3 = 10 t3=10t

t = 0.3 st=0.3s

now the distance moved by the lower drop in this time

d_1 = \frac{1]{2}gt^2

d_1 = 0.45 md

1

=0.45m

now we know that it drop 4 drops per sec

so the time interval between two drops is 0.25 s

so here we can say that the distance covered by 2nd drop in t = 0.3 - 0.25 s

d_2 = \frac{1}{2}(10)(0.05)^2d

2

=

2

1

(10)(0.05)

2

d_2 = 0.0125 md

2

=0.0125m

so the distance between two drops is given as

d = 0.45 - 0.0125d=0.45−0.0125

d = 0.4375 md=0.4375m

Answer 2

Answer:

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Explanation: