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3 years ago

6

Six artificial satellites circle a space station at a constant speed. The mass m of each satellite, distance L from the space st

ation, and the speed v of each satellite are listed below. The satellites fire rockets that provide the force needed to maintain a circular orbit around the space station. The gravitational force is negligible.
Rank each satellite based on its acceleration. Rank from largest to smallest.

A) m= 200 kg, L= 5000 m, v= 160 m/s

B) m= 100 kg, L= 2500 m, v= 160 m/s

C) m= 400 kg, L= 2500 m, v= 80 m/s

D) m= 800 kg, L= 10000 m, v= 40 m/s

E) m= 200 kg, L= 5000 m, v= 120 m/s

F) m= 300 kg, L= 10000 m, v= 80 m/s

2 answers:

Soloha48 [4]3 years ago

5 0

Explanation:

Below is an attachment containing the solution.

damaskus [11]3 years ago

3 0

Answer:

The answer to the question is;

Based on their acceleration the rank of the satellites from largest to smallest is.

B >→ A >→ E >→ C >→ F >→ D.

Explanation:

Acceleration is given by $\frac{v^2}{r}$

Therefore the acceleration for each of the satellite is given by

Satellite A) $\frac{160^2}{5000}$ = 5.12 m/s²

Satellite B) $\frac{160^2}{2500}$ = 10.24 m/s²

Satellite C) $\frac{80^2}{2500}$ = 2.56 m/s²

Satellite D) $\frac{40^2}{10000}$ = 0.16 m/s²

Satellite E) $\frac{120^2}{5000}$ = 2.88 m/s²

Satellite F) $\frac{80^2}{10000}$ = 0.64 m/s²

Therefore in order of decreasing acceleration, from largest to smallest we have

Satellite B) > Satellite A) >Satellite E) >Satellite C)>Satellite F)>Satellite D).

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Explanation:

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This means that the electric potential is constant, so we can write:

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i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

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$V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} ln\frac{b}{r}$

The electric field is just the derivative of the electric potential:

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$E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}$

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