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Firdavs [7]
1 year ago
13

during a lunar mission, it is necessary to increase the speed of a spacecraft by 2.2 m/s when it is moving at 400 m/s relative t

o the moon. the speed of the exhaust products from the rocket engine is 1000 m/s relative to the spacecraft. what fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?
Physics
1 answer:
wel1 year ago
3 0

The initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

<h3>What fraction of the initial mass of the spacecraft?</h3>

Increase the speed: Vf-Vi = 2.2 m/s

Speed of aircraft: Vr = 400 m/s

Speed of ejected products: Vrel = 1000 m/s

The answer is:

V_f - V_i = V_{rel} log_{e} (\frac{mi}{mf})\\\\2.2 = 1000 log_{e} (\frac{mi}{mf})\\\\ log_{e} (\frac{mi}{mf} ) = \frac{2.2}{1000} \\\\ log_{e} (\frac{mi}{mf} ) = 0.0022

\frac{mi}{mf} = e^{0,0022} \\\\\frac{mf}{mi} = e^{-0,0022} \\\\\frac{mi-mf}{mi} = 1 - e^{-0,0022} = 0,00219

So, the initial mass fraction of the spacecraft that must be burned and ejected to achieve an increase in speed is 0,00219 m/s

Learn more about spaceship speed fraction brainly.com/question/28256735

#SPJ4

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Calculate the acceleration of a 270000-kg jumbo jet just before takeoff when the thrust on the aircraft is 160000 N .
Radda [10]

Answer:

<h3>The answer is 0.59 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{160000}{270000}  =  \frac{16}{27}  \\  = 0.592592...

We have the final answer as

<h3>0.59 m/s²</h3>

Hope this helps you

7 0
3 years ago
Help me asap its due today
Margaret [11]

Answer:

3.54* 10^{22} N

Explanation:

Using the formula you gave:

F_g = \frac{6.67*10^{-11}*2.0*10^{30}*5.97^{24}  }{(1.5*10^{11})^2 }

3 0
3 years ago
An 85.0 kg fisherman jumps from a dock at a speed of 4.30 m/s onto their 135.0 kg boat. If the boat was at rest to begin but mov
jeka94

Answer:

Final speed of boat + man is 1.66 m/s

Explanation:

As we know that there is no friction on the system or there is no external force on this system

So here we can use momentum conservation here

mv = (m + M)v_f

so we have

m = 85 kg

M = 135 kg

v = 4.30 m/s

now we have

85 \times 4.30 = (85 + 135) v

v = 1.66 m/s

4 0
3 years ago
How large is the acceleration of a 25 kg mass with a net force of 75 N applied horizontally to it?
Sveta_85 [38]

Answer:

Explanation:

F = ma

a = F/m

a = 75/25

a = 3 m/s²

8 0
2 years ago
A bowling ball is far from uniform. Lightweight bowling balls are made of a relatively low-density core surrounded by a thin she
tester [92]

Answer:

a)  I = 1,75 10-² kg m²  and b)  I = 1.49 10⁻² kg m²

Explanation:

The expression for the moment of inertia is

    I = ∫ r² dm

The moment of inertia is a scalar by which an additive magnitude, we can add the moments of inertia of each part of the system, taking into account the axis of rotation.

    I = I core + I shell

The moment of inertia of a solid sphere is

    I sphere = 2/5 MR²

The moment of inertia of a thin spherical shell is

    I shell = 2/3 M R²

a) Let's apply to our system, first to the core of weight 1.6 kg and diameter 0.196m, the radius is half the diameter

     R = d / 2

     R= 0.196 m / 2 = 0.098 m

     I core = 2/5 1.6 0.098²

     I core = 6.147 10-3 kg m²

Let's calculate the moment of inertia of the shell of mass 1.6 kg with a diameter of 0.206 m

    R = 0.206 / 2

    R = 0.103 m

    I shell = 2/3 1.6 0.103²

    I shell = 1,132 10-2 kg m²

The moment of inertia of the ball is the sum of these moments of inertia,

    I = I core + I shell

    I = 6,147 10⁻³ + 1,132 10⁻² = 6,147 10⁻³ + 11.32 10⁻³

    I = 17.47 10⁻³ kg m²

    I = 1,747 10-² kg m²

b) Now the ball is report with mass 3.2kg and diameter 0.216 m

    R = 0.216 / 2

    R = 0.108 m

It is a uniform sphere

    I = 2/5 M R²

    I = 2/5 3.2 0.108²

    I = 1.49 10⁻² kg m²

7 0
3 years ago
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