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Blababa [14]
2 years ago
9

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane

t's surface. The period of revolution of the satellite around the planet is T 1.15 hours. What is the average density of the planet?
Physics
1 answer:
lubasha [3.4K]2 years ago
8 0

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

T = \frac{2\pi (r + h)}{v}

T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

M = 4.54 \times 10^{23} kg

Now the density is given as

\rho = \frac{M}{\frac{4}{3}\pi r^3}

\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

\rho = 12580.7 kg/m^3

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A 2kg book is held against a vertical wall. The coefficient of friction is 0.45. What is the minimum force that must be applied
Vika [28.1K]

We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that  the minimum force that must be applied on the <em>book is</em>

  • F=44N

From the question we are told

A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal

Generally the equation for the  Force  is mathematically given as

F=\frac{mg}{\mu}\\\\F=\frac{2*9.8}{0.45}\\\\

F=44N

Therefore

the minimum force that must be applied on the <em>book is</em>

F=44N

For more information on this visit

brainly.com/question/23379286

8 0
2 years ago
PY85
aliya0001 [1]

Answer:

460 g

Explanation:

Heat lost by the warm water = heat gained by the cold water

-mCΔT = mCΔT

-m (4.184 J/g/K) (37°C − 85°C) = (1000 g) (4.184 J/g/K) (37°C − 15°C)

-m (37°C − 85°C) = (1000 g) (37°C − 15°C)

-m (-48°C) = (1000 g) (22°C)

m = 458 g

Rounded to two significant figures, you need a mass of 460 g of water.

3 0
2 years ago
A 90 kg body is taken to a planet where the acceleration due to
Serggg [28]

Answer:

2250N

Explanation:

W= mg,

where W= weight

m= mass

g= acceleration due to gravity

Given that the body is 90kg, m= 90kg.

Acceleration due to gravity of planet

= 2.5(10)

= 25 m/s²

Weight of body on planet

= 90(25)

= 2250N

*Mass is the amount of matter an object has and is constant (same on earth and the planet).

6 0
3 years ago
Read 2 more answers
Is a cinder cone volcanoe constructive or deconstructive
Ksivusya [100]
Cinder cone volcanoes can be associated with either constructive or destructive margins.<span>
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6 0
3 years ago
The displacement of a car is a function of time as follows: x(t)=25+3.0t², with x is in meters. Find the average velocity betwee
aniked [119]

Answer: 15m/s

Explanation: <u>Average</u> <u>Velocity</u> is vector describing the total displacement of an object and the time taken to change its position. It is represented as:

v=\frac{\Delta x}{\Delta t}

At t₁ = 1.0s, displacement x₁ is:

x(1)=25+3(1)^{2}

x(1) = 28

At t₂ = 4.0s:

x(4)=25+3(4)^{2}

x(4) = 73

Then, average speed is

v=\frac{73-28}{4-1}

v = 15

The average velocity of a car between t₁ = 1s and t₂ = 4s is 15m/s

5 0
2 years ago
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