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Anuta_ua [19.1K]
3 years ago
12

Two satellites are in circular orbits around Earth. Satellite A has speed vA. Satellite B has an orbital radius nine times that

of satellite A. What is the speed of Satellite B?
A. vA.9
B. vA/3
C. 3vA
D. 9vA
Physics
1 answer:
zaharov [31]3 years ago
5 0

Answer:

option B

Explanation:

given,

Satellite B has an orbital radius nine times that of satellite A.

R' = 9 R

now, orbital velocity of the satellite A

v_a =\sqrt{\dfrac{GM}{R}}........(1)

now, orbital velocity of satellite B

v_b=\sqrt{\dfrac{GM}{R'}}

v_b=\sqrt{\dfrac{GM}{9R}}

v_b=\dfrac{1}{3}\sqrt{\dfrac{GM}{R}}

from equation 1

v_b=\dfrac{v_a}{3}

hence, the correct answer is option B

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Which has a higher acceleration: a 10-kg object acted upon with a net force of 20 N or an 18-kg object acted on by a net force o
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the sole of a tennis shoe has a surface area of 0.0290 m^2. if it is worn by a 65.0 kg person, what pressure does the shoe exert
AURORKA [14]

Answer: 21965.517 Pa

Explanation:

Pressure P is the force F exerted by a gas, a liquid or a solid on a surface (or area) A, its unit is Pascal Pa which is equal to N/m^{2} and its formula is:  

P=\frac{F}{A} (1)

In this case we have the surface of a sole of a tennis shoe:

A=0.0290 m^{2} (2)

And the mass m of the person who wears it:

m=65 kg

On the other hand, we know the weight is the force  F the Earth exerts on people and objects due gravity g :

F=m.g=(65 kg)(9.8m/^{2})

F=637N (3)

Substituting (2) and (3) in (1):

P=\frac{637N}{0.0290 m^{2}} (4)

Finally:

P=21965.517 Pa This is the pressure the shoe exert on the ground

5 0
3 years ago
At high speeds, a particular automobile is capable of an acceleration of about 0.540 m/s^2. At this rate, how long (in seconds)
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Answer:

t = 6.68 seconds

Explanation:

The acceleration of the automobile, a=0.54\ m/s^2

Initial speed of the automobile, u = 91 km/hr = 25.27 m/s

Final speed of the automobile, v = 104 km/hr = 28.88 m/s

Let t is the time taken to accelerate from u to v. It can be calculated as the following formula as :

t=\dfrac{v-u}{a}

t=\dfrac{28.88-25.27}{0.54}

t = 6.68 seconds

So, the time taken by the automobile to accelerate from u to v is 6.68 seconds. Hence, this is the required solution.

5 0
3 years ago
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