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Gnom [1K]
3 years ago
12

A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I

f the magnitude of the gravitational acceleration on the surface of the planet is 20.0 m/s2, what is the radius of the planet?
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
8 0

Answer:

Explanation:

Expression for times period of a satellite can be given as follows

Time period T = 1.8 x 60 x 60

= 6480

T² = \frac{4\times \pi^2\times r^3}{GM} where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.

6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM

GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

20 = 3.96 X 10¹⁴ / R²

R² = 3.96 X 10¹⁴ / 20

= 1.98 x 10¹³ m

R= 4.45 x 10⁶ m

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Deep in the interiors of the giant planets, water is still a liquid even though the temperatures are tens of thousands of degree
Nataliya [291]

Answer:

High pressure inside the giant planet

Explanation:

As we move in the interior of the giant planet, the pressure and temperature in the interior of the planet increases. Since, the giant planets have hardly any solid surface and thus they are mostly constituted of atmosphere.

Also, the gravitational forces keep even the lightest of the matter bound in it contributing to the large mass of the planet.

If we look at the order of the magnitude of the temperature of these giant planets than nothing should be able to stay in liquid form but as the depth of the planet increases with the increase in temperature, pressure also increases which keeps the particle of the matter in compressed form.

Thus even at such high order of magnitude water is still found in liquid state in the interior of the planet.

7 0
3 years ago
f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t
inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

h(t)=35t-0.83t^2

We know that the rate of change of displacement is equal to the velocity of an object.

v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t

Velocity of the arrow after 3 seconds will be :

v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

7 0
3 years ago
The density of mobile electrons in copper metal is 8.4 × 1028 m-3. Suppose that i= 4.4 × 1018 electrons/s are drifting through a
neonofarm [45]

Answer: 405.3 minutes

Explanation: In order to explain this problem we have to use the following:

Fisrtly we calculate the volume of the wire, this is given by:

Vwire=π*r^2*L where r and L are the radius and L the length of teh wire, respectively.

Vwire=π*1.25*10^-3*0.26=1.27*10^-6 m^3

then the number of the total electrons in tthe wire volume is given by;

n° electrons in the wire=ρ*Vwire=8.4*10^28*1.27*10^-6 m^3=1.07 *10^23

Finally, considering the current in the wire equal to 4.4*10^18 electrons/s

the time consuming to extract all the electrons from the wire is given by:

t= total electrons in the wire/ current=1.067*10^23/4.4*10^18=24,318 s

equivalent to 405.3 minutes

4 0
3 years ago
If a data point is way off the trend line which of the following will not help resolve the problem
S_A_V [24]
Need more details to answer
3 0
3 years ago
How much Power is needed to lift a 200 N block
vova2212 [387]

Answer:

10 W

Explanation:

P = F x v

F = 200N

v = x2 - x1 / t

  = 6 - 3 / 60

  = 3 / 60

  = 1 / 20

  = 0.05 m/s

P = 200 x 1/20

  = 10 W

4 0
2 years ago
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