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Vilka [71]
3 years ago
8

A 5000-kg spacecraft is in a circular orbit 2000 km above the surface of mars. how much work must the spacecraft engines perform

to move the spacecraft to a circular orbit that is 4000 km above the surface?
Physics
1 answer:
Elden [556K]3 years ago
8 0
The spacecraft is orbiting at 2000 Km. The engine has to put it on 4000 km altitude, therefore the engine has to push it into another 2000 km, so that it will attain the 4000 km altitude.

Given. Gravitational pull of planet Mars g = 3.71 m/s²

            Mass of the spacecraft m = 5000 kg

Formula: W = fd   F = mg    

               W = mgh

               W = (5000 Kg)(3. 71 m/s²)(2 x 10⁶m)

               W = 3.71 x 10¹⁰ J

             
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Please help<br> types of stimuli the sensory organs receive.
Alborosie

Answer: Photoreceptor cell, Chemoreceptor, Thermoreceptor, Mechanoreceptor

Hope this helps(:

Explanation:

3 0
2 years ago
Which statement best describes a characteristic of gases?
salantis [7]
Assumes the shape and volume of its container 
<span>particles can move past one another</span>
8 0
3 years ago
A 70 kg student stands on top of a 5.0 m platform diving board . how much gravitational potential energy does the student have?
Marianna [84]

Answer:

a. P.E = 3430Joules.

b. Workdone = 3430Nm

Explanation:

<u>Given the following data;</u>

Mass = 70kg

Distance = 5m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy = mgh

P.E = 70*9.8*5

<em>P.E = 3430J</em>

b. To find the workdone;

Workdone = force * distance

But force = mass * acceleration

Force = 70*9.8

Force = 686 Newton.

Workdone = 686 * 5

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6 0
2 years ago
Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the
BlackZzzverrR [31]

relation between potential difference and electric field is given as

E . d = \Delta V

so here we know that

d = 3 cm

\Delta V = 30 V

E \times 0.03 = 30

E = 1000 N/C

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

5 0
3 years ago
The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c
Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0
2 years ago
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