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Nataly_w [17]
2 years ago
15

A spaceprobe in outer space is flying with a constant speed of 1.795 km/s. The probe has a payload of 1635.0 kg and it carries 4

092.0 kg of rocket fuel. The rocket engines of the probe are capable of expelling propellant at a speed of 4.161 km/s. Then the rocket engines are fired up. How fast will the spaceprobe travel when all the rocket fuel is used up?
Physics
1 answer:
erma4kov [3.2K]2 years ago
7 0

The speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

<h3>What is the law of conservation of linear momentum?</h3>

According to the law of conservation of linear momentum before the collision is equal to the momentum after the collision. These laws state how momentum gets conserved.

Unit conversion;

1 km/sec = 1000 m/sec

Given data;

Spaceprobe speed  = 1.795 km/s = 1795 m /sec

Probe mass = 635.0 kg

Fuel mass = 4092.0 kg

Expelled propellent velocity = 4.161 km/s = 41461 m/sec

From the momentum conservation principle;

\rm P_i = P_f \\\\ (m_p+m_f)v_i = m_pV - m_fv_p \\\\ V = \frac{(635+4092)1795+4092 \times  41461}{635} \\\\ V = 280540.7 \ m/sec \\\\ V = 28.05 m/sec

Hence, the speed by which the spaceprobe travels when all the rocket fuel is used up will be 29.262 m/sec.

To learn more about the law of conservation of momentum refer to:

brainly.com/question/1113396

#SPJ1

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mesa

Explanation:



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5 0
3 years ago
A stone is thrown with an initial speed of 11.5 m/s at an angle of 50.0 above the horizontal from the top of a 30.0-m-tall build
rewona [7]

Answer:

The magnitude of the horizontal displacement of the rock is 7.39 m/s.

Explanation:

Given that,

Initial speed = 11.5 m/s

Angle = 50.0

Height = 30.0 m

We need to calculate the horizontal displacement of the rock

Using formula of horizontal component

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=11.5\times\cos50

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The escape velocity of any object from Earth is 11.2 km/s. (a) Express this speed in m/s and km/h. (b) At what temperature would
natima [27]

Answer:

a ) 11.1 *10^3 m/s = 39.96 Km/h

b) T_{o2} =1.58*10^5 K

Explanation:

a)v_{es} =v_{rms}= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h

b)

M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol

gas constant R = 8.31 j/mol.K

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So, v_{rms,o2} =\sqrt{ \frac{3RT_{o2}}{M_{o2}}}

multiply each side by M_{o2}, so we have

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solving for temperature T_{o2}

T_{o2} = \frac{v_{rms,o2}^2 *M_{o2}}{3R}

In the question given,v_{rms} =v_{es}

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T_{o2} =1.58*10^5 K

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