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Elodia [21]
3 years ago
11

A satellite orbiting Earth has a tangential velocity of 5000 m/s. Earth’s mass is 6 × 1024 kg and its radius is 6.4 × 106 m.

Physics
2 answers:
Bumek [7]3 years ago
7 0

Answer:

The answer is 9,600,000 on edgen

Aloiza [94]3 years ago
4 0
When a satellite is moving around the Earth's orbit, two equal forces are acting on it. The centripetal and the centrifugal force. The centripetal force is the force that attracts the object toward the center of the axis of rotation. The opposite force is the centrifugal force. It draws the object away from the center. When these forces are equal, the satellite uniformly rotates along the orbit.

Centripetal force = Centrifugal force
Mass of satellite * centripetal acceleration = Mass of satellite * centrifugal acceleration
Centripetal acceleration = Centrifugal acceleration

\frac{ M_{E}*G }{  ( r_{E} )^{2}  } =ω^2r

where

M_{E} = mass of earth
G = gravitational constant = 6.6742 x 10-11<span> m</span>3<span> s</span>-2<span> kg</span><span>-1
</span>r_{E} = radius of earth
ω = angular velocity
<span>r = radius of orbit

To convert to angular velocity:

</span>Tangential velocity = rω
ω = 5000/r

Then,

\frac{ (6 \ x \  10^{24}) *(6.6742 \ x \ 10^{-11} ) }{ ( 6.4 \ x \  10^{6})^{2} }= ( \frac{5000}{r} )^{2} r

r = 2557110.465 m

Therefore, the distance of the centers of the earth and the satellite is 2.6 x 10^6 m.
<span>
</span>
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A basketball player throws a chall -1 kg up with an initial speed of his hand at shoulder height = 2.15 m Le gravitational poten
Talja [164]

Complete Question:

A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level

a)  Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.

b) What is the height, hn in meters?

Answer:

a) Energy = mghₙ

b) Height, hₙ = 5.02 m

Explanation:

a) Total energy in terms of maximum height

Let maximum height be hₙ

At maximum height, velocity, V=0

Total mechanical energy , E = mgh + 1/2 mV^2

Since V=0 at maximum height, the total energy in terms of maximum height becomes

Energy = mghₙ

b) Height,  hₙ in meters

mghₙ = mgh + 1/2 mV^2

mghₙ = m(gh + 1/2 V^2)

Divide both sides by mg

hₙ = h + 0.5 (V^2)/g

h = 2.15m

g = 9.8 m/s^2

V = 7.5 m/s

hₙ = 2.15 + 0.5(7.5^2)/9.8

hₙ = 2.15 + 2.87

hₙ = 5.02 m

6 0
3 years ago
As you stand by the side of the road, a car approaches you at a constant speed, sounding its horn, and you hear a frequency of 8
kap26 [50]

Answer: velocity of the car is 113.33m/s

Explanation:

From Doppler effect,

in the case which the source is moving towards the observer at rest

f2 = v/(v-vs) *f1

where f2 is the final observed frequency

f1 is the initial observed frequency

v = 340m/s (speed of sound in air)

vs = velocity of the source of sound.

rearranging the above equation

f2*(v - vs) = f1* v

vs = (f1* v/f2) - v

but f1 = 80Hz

f2 = 60Hz

v = 340m/s

substituting,

vs = (80 x 340)/60 - 340

vs = 453.33 - 340

vs = 113.33m/s

velocity of the car is 113.33m/s

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3 years ago
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Answer:

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4 0
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vlabodo [156]
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3 0
3 years ago
To view an interactive solution to a problem that is similar to this one, select Interactive Solution 7.24. A 0.0129-kg bullet i
ipn [44]

Answer:

t=0.42s

Explanation:

Here you have an inelastic collision. By the conservation of the momentum you have:

m_1v_1+m_2v_2=(m_1+m_2)v

m1: mass of the bullet

m2: wooden block mass

v1: velocity of the bullet

v2: velocity of the wooden block

v: velocity of bullet and wooden block after the collision.

By noticing that after the collision, both objects reach the same height from where the wooden block was dropped, you can assume that v is equal to the negative of v2. In other words:

m_1(-v_1)+m_2v_2=(m1+m2)(-v2)

Where you assumed that the negative direction is upward. By replacing and doing v2 the subject of the formula you get:

-(0.0129kg)(767m/s)+(1.17kg)v_2=(1.1829kg)(-v_2)\\\\v_2=4.20m/s

Now, with this information you can use the equation for the final speed of an accelerated motion and doing t the subject of the formula. IN other words:

v_2=v_o+gt\\\\t=\frac{v_2-v_o}{g}=\frac{4.2m/s-0m/s}{9.8m/s^2}=0.42s

hence, the time is t=0.42 s

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3 years ago
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