An elevator moves downward in a tall building at a constant speed of 5.05 m/s. Exactly 5.75 s after the top of the elevator car passes a bolt loosely attached to the wall of the elevator shaft, the bolt falls from rest. (a) At what time does the bolt hit the top of the still-descending elevator? (Assume the bolt is dropped at t = 0 s.)

(b) Estimate the highest floor from which the bolt can fall if the elevator reaches the ground floor before the bolt hits the top of the elevator. (Assume 1 floor ≌ 3 m.)

Explanation:

The elevator is moving with uniform speed and bolt will fall with acceleration due to gravity g .

Displacement of elevator when bolt starts falling

= 5.05 x 5.75 = 29 m

Let after time t bolt reaches the elevator

in time t , displacement of bolt = 1/2 g t²

= .5 x 9.8 x t² = 4.9 t²

displacement of elevator = 5.05 t

position of elevator = 29 + 5.05 t

According to question ,

29 + 5.05 t = 4.9 t²

4.9 t² - 5.05 t - 29 = 0

t = 3 s approx .

b )

Distance covered by the bolt

= 1 /2 g t²

= .5 x 9.8 x 9

= 44.1 m

Bolt must have to fall from a height of 44.1 m

or in terms of floor it is

44.1 / 3 = 15 th floor .

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.