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zvonat [6]
3 years ago
5

I’m stuck in B and D. which one is it?

Physics
2 answers:
Varvara68 [4.7K]3 years ago
8 0

<u>A</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>:</u>

Option b is correct - Rocky, nearly continuous, found on earth surface.

The reason is <u>i</u><u>t</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>r</u><u>o</u><u>c</u><u>k</u><u>y</u><u> </u><u>s</u><u>u</u><u>r</u><u>f</u><u>a</u><u>c</u><u>e</u><u> </u><u>a</u><u>r</u><u>e</u><u>a</u><u> </u><u>c</u><u>o</u><u>m</u><u>p</u><u>o</u><u>s</u><u>e</u><u>d</u><u> </u><u>o</u><u>f</u><u> </u><u>c</u><u>r</u><u>u</u><u>s</u><u>t</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>u</u><u>p</u><u>p</u><u>e</u><u>r</u><u> </u><u>m</u><u>a</u><u>n</u><u>t</u><u>l</u><u>e</u><u> </u><u>p</u><u>a</u><u>r</u><u>t</u><u>.</u><u> </u><u>B</u><u>a</u><u>s</u><u>i</u><u>c</u><u>a</u><u>l</u><u>l</u><u>y</u><u>,</u><u> </u><u>i</u><u>t</u><u> </u><u>i</u><u>s</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>o</u><u>u</u><u>t</u><u>e</u><u>r</u><u> </u><u>s</u><u>h</u><u>e</u><u>l</u><u>l</u><u> </u><u>o</u><u>f</u><u> </u><u>e</u><u>a</u><u>r</u><u>t</u><u>h</u><u>.</u>

tatuchka [14]3 years ago
6 0
It is b! the lithosphere is the crust and upper mantle
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A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.
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a) The acceleration of gravity is 4.96 m/s^2

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

a)

The acceleration of gravity for an object near a planet is given by

g=\frac{GM}{(R+h)^2}

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2

b)

The critical velocity for a satellite orbiting around a planet is given by

v=\sqrt{\frac{GM}{R+h}}

where we have again:

M=5.98\cdot 10^{24} kg (mass of the Earth)

R=6.37\cdot 10^6 m (Earth's radius)

h=2.60\cdot 10^6 m (altitude of the satellite)

Substituting,

v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s

Converting into km/h,

v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h

c)

The period of the orbit is given by the circumference of the orbit divided by the velocity:

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

where

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

v = 6668 m/s

Substituting,

T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s

d)

One day consists of:

t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

n=\frac{t}{T}=\frac{86400}{8452}=10.2

e)

The escape velocity for an object in the gravitational field of a planet is given by

v=\sqrt{\frac{2GM}{R+h}}

where here we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=2.60\cdot 10^6 m

Substituting, we find

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s

f)

We can apply again the formula to find the escape velocity for the rocket:

v=\sqrt{\frac{2GM}{R+h}}

Where this time we have:

M=5.98\cdot 10^{24} kg

R=6.37\cdot 10^6 m

h=0, because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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