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myrzilka [38]
3 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?
Physics
2 answers:
katovenus [111]3 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]3 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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san4es73 [151]

Answer:

49.7^{\circ}

Explanation:

We can solve the problem by using Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1 = 1.33 is the index of refraction of the first medium (water)

n2 = 1.00 is the index of refraction of the second medium (air)

\theta_1 = 35^{\circ} is the angle of incidence of the wave in water

\theta_2 is the angle of refraction of the wave in air

Solving for the angle of refraction,

sin \theta_2 = \frac{n_1 sin \theta_1}{n_2}=\frac{(1.33) sin 35^{\circ}}{1.00}=0.763\\\theta_2 = sin^{-1} (0.763)=49.7^{\circ}

6 0
3 years ago
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