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4 years ago

13

Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.

This process takes 10 s. During the time interval what average force is applied to the rocket?

2 answers:

katovenus [111]4 years ago

4 0

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

M v + m v' = 0

5000 x v - 10 x 4000 = 0

5000 v = 40000

v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

$F = \dfrac{m(v_i-v_f)}{\Delta t}$

$F = \dfrac{5000(8-0)}{10}$

F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

Yuri [45]4 years ago

3 0

Answer:

4000 N

Explanation:

mass of fuel, m = 10 kg

velocity of ejected fuel, v = 4000 m/s

time, t = 10 s

force, F = velocity x mass of fuel burning per second

F = 4000 x 10 / 10 = 4000 N

Thus, the force is 4000 N.

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4 years ago

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a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista

ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$ (1)

Where:

$x_{f}$ : is the final position in the horizontal direction = 80 m

$x_{0}$ : is the initial position in the horizontal direction = 0

$v_{0_{x}}$ : is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

$t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s$

b) The height of the hill is given by:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final position in the vertical direction = 0

$y_{0}$ : is the initial position in the vertical direction =?

$v_{0_{y}}$ : is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

$y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m$

I hope it helps you!

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3 years ago

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A rocket has initail mass M begins to move from space , with an exhaust constant speed . Find the mass of the rocket while has t

irina1246 [14]

Answer:

$m=\frac{m_{0}}{e}$

Explanation:

Equation of the rocket is,

$m\frac{dv}{dt} =F-v'\frac{dm}{dt}$

Here, v' is the relative velocity of rocket.

In space F is zero.

So,

$m\frac{dv}{dt} =-v'\frac{dm}{dt}\\dv=-v'\frac{dm}{m} \\v=-v'ln\frac{m}{m_{0} }$

Now the momentum can be obtained by multiply by m on both sides.

$P=-v'mln\frac{m}{m_{0} }$

Now for maxima, $\frac{dP}{dm}=0$

$-v'ln\frac{m}{m_{0} }-v'm\frac{m_{0}}{m }m_{0=0$

Now,

$ln(\frac{m}{m_{0} } )=-1\\\frac{m}{m_{0} }=\frac{1}{e} \\m=\frac{m_{0}}{e}$

Therefore, the mass of the rocket while having maximum momentum is $\frac{m_{0}}{e}$

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4 years ago