Question

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Answer 1

a. The velocity t = $v = Ce_{n} (\frac{mo}{mo - kt} ) - gt$

b. v60 = 7164

How to solve for the velocity

mdv/dt = ck - mg

dv/dt = ck/m - mg/m

= ck/m - g

dv = $(\frac{ck}{Mo-Kt} -g)dv$

Integrate the two sides of the equation to get

v $-\frac{ck}{k} e_{n} (Mo- kt)-gt+c$

$v = Ce_{n} (\frac{mo}{mo - kt} ) - gt$

b. fuel accounts for 55% of the mass

So final mass after fuel is burned out is = 0.45

c=2500

g=9.8

t=60

v = -2500ln0.45 - 9.8 x 60

= 7752 - 588

= 7164

Complete question

A rocket, fired from rest at time t = 0, has an initial mass of m0 (including its fuel). Assuming that the fuel is consumed at a constant rate k, the mass m of the rocket, while fuel is being burned, will be given by m0 - kt. It can be shown that if air resistance is neglected and the fuel gases are expelled at a constant speed c relative to the rocket, then the velocity of the rocket will satisfy the equation where g is the acceleration due to gravity.

dv dt m =ck - mg

(a) Find v(t) keeping in mind that the mass m is a function of t.

v(t) =

m/sec

(b) Suppose that the fuel accounts for 55% of the initial mass of the rocket and that all of the fuel is consumed at 60 s. Find the velocity of the rocket in meters per second at the instant the fuel is exhausted. [Note: Take g = 9.8 m/s² and c = 2500 m/s.]

v(60) =

m/sec [Round to nearest whole number]

Raed more on velocity here

brainly.com/question/25749514

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