A two-stage rocket is traveling at 1210m/s with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 40m/s relative to the second stage after the explosion. The first stage is three times as massive as the second stage.

What is the speed of the second stage after the separation?

R

Explanation:

Given that,

Two stage rocket traveling at

V = 1210 m/s with respect to earth

First stage

When fuel Is run out , explosive bolts releases and push rocket backward at speed is

V1 = 40m/s relative to second stage

Therefore

V1 = 40 - V2

The first stage is 3 times as massive as the second stage

I.e Mass of first stage is 3 times the second stage

Let Mass of second stage be

M2 = M

Then, M1 = 3M

Velocity of second stage V2?

Applying conservation of linear momentum

Momentum before explosion = momentum after explosion

Momentum p=mv

Then,

(M1+M2)V = —M1•V1 + M2•V2

(3M+M)•1210 = —3M•(40-V2) +M•V2

4M × 1210 = —120M + 3M•V2 + MV2

4840M = —120M + 4M•V2

4840M + 120M = 4M•V2

4960M = 4M•V2

Then, V2 = 4960M / 4M

V2 = 1240 m/s

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.