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NISA [10]
3 years ago
15

) A satellite of mass m has an orbital period T when it is in a circular orbit of radius R around the earth. If the satellite in

stead had mass 4m, its orbital period would be A) T. B) 2T. C) 4T. D) T/4. E) T/2
Physics
1 answer:
Mrrafil [7]3 years ago
4 0

Answer:

A) T.

Explanation:

Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:

T=\frac{2\pi R^{\frac{3}{2}}}{\sqrt{GM}}  

So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.

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If you pour different liquids into a graduated cylinder, which layer would be at the bottom?
klasskru [66]

the one that is more dense!

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3 years ago
A flowerpot falls from the ledge of an apartment building. A person in an apartment below, coincidentally holding a stopwatch, n
valkas [14]

Answer

given,

height of window = 4 m

time taken to travel = 1 s

acceleration due to gravity = 9.8 m/s²

s = ut + \dfrac{1}{2}at^2

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u = \dfrac{4 - \dfrac{1}{2}\times 9.8\times 1^2}{1}

u = -0.905 m/s

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now,

v² = u² + 2 a s

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6 0
3 years ago
Which choice correctly describes what happens during heating?
Masja [62]

Answer:

Option 4

Explanation:

During heating actually heat transfer takes place from a body at higher temperature to a body at lower temperature and the heat transfer takes place until both attain the same temperature  

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So in this case assuming that both the systems have same mass then the energy will leave the system with greater thermal energy and go into the system with less thermal energy as the system with greater thermal energy in this case will be at higher temperature and we are considering this assumption because thermal energy not only depends on temperature but also depends on mass of the system

7 0
3 years ago
Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m
umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s

v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads

Now, we just use the formula:

w_f^2-w_o^2=2\alpha*x

\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2

6 0
3 years ago
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