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zavuch27 [327]
3 years ago
9

A spaceprobe in outer space is flying with a constant speed of 1.530 km/s. The probe has a payload of 1363.0 kg and it carries 3

486.0 kg of rocket fuel. The rocket engines of the probe are capable of expelling propellant at a speed of 3.795 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:
abruzzese [7]3 years ago
3 0

Answer:

6.33 km/s

Explanation:

Given that :

A spaceprobe in outer space is flying with a constant speed v_i =  1.530 km/s.

The probe has a payload =  1363.0 kg

which carries 3486.0 kg of rocket fuel.

Exhaust speed = 3.795 km/s

How fast will the spaceprobe travel when all the rocket fuel is used up?

As we know that the rate of change of spaceprobe momentum is equal to the thrust of the rocket.

Then;

m \frac{dv}{dt} = -v_{ex} \frac{dm}{dt}

where;

v_{et  = exhaust speed

dv = -v_{ex}\frac{dm}{m}

Taking the integral of the above expression; we have:

v_f -v_i = - v_{ex}In m|^{m_f}_{m_o}

v_f -v_i = - v_{ex}In \frac{m_o}{m_f}

v_f  = v_i +  v_{ex}In \frac{m_o}{m_f}

v_f  =1.530 +  3.795 In (\frac{1363+3486}{1363} )

= 6.33 km/s

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If the value of static friction for a couch is 400N, what could be a possible value of its kinetic friction? *
Dafna11 [192]

Answer:

kinetic friction may be greater than 400 N or smaller than 400 N

Explanation:

As we know that maximum value of static friction on the rough surface is known as limiting friction and the formula of this limiting friction is known as

F_s = \mu_s N

now when object is sliding on the rough surface then the friction force on that surface is known as kinetic friction and the formula of kinetic friction is known as

F_k = \mu_k N

now we know that

\mu_k < \mu_s

so here value of limiting static friction force is always more than kinetic friction

also we know that

initially when body is at rest then static friction value will lie from 0 N to maximum limiting friction

and hence kinetic friction may be greater than static friction or if the static friction is maximum limiting friction then kinetic friction is smaller than static friction

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5 0
3 years ago
A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also state
balu736 [363]

To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as

I = F*t

Where,

F= Force

t= time

At the same time the moment can be described as a function of mass and velocity, that is

P = m\Delta v \rightarrow P=m(v_1-v_2)

Where,

m = mass

v = Velocity

From equilibrium the impulse is equal to the momentum, therefore

I = p

Ft = m(v_1-v_2)

PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be

p=m(v_1-v_2)

p = 75*0.15

p = 1125Kg\cdot m/s

Therefore the magnitude of the person's impulse is 1125Kg.m/s

PART B) From the equation obtained previously we have that the Force would be:

Ft = m(v_1-v_2)

F(0.025)= 1125

F= 45000N

Therefore the magnitude of the average force the airbag exerts on the person is 45000N

6 0
3 years ago
How is the surface area of an average person is 2m^2 in the chapter pressure
11Alexandr11 [23.1K]

Answer:

We have to show surface area =2m^2,with few conditions that is by considering Force =200000\ N and Pressure =100000\ Pa to be respectively.

Explanation:

The atmospheric pressure is =10^{5}\ Pa on Earth's surface.

The magnitude of the force exerted on a person by the atmosphere is =2\times 10^{5}\N (or)\ 200kN\ (or)\ 20\ ton.

Now to calculate surface area we can find it from pressure=\frac{force}{area} and re-arranging it to.

area=\frac{force}{pressure}

So plugging the values,

Surface area =\frac{20000}{10000}=2\ m^{2}

Hence from the above calculations we can say that surface area is 2m^2.

So the surface area of an average person can be said to have 2m^2, using the concept of pressure and force.

3 0
3 years ago
Read 2 more answers
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