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Kryger [21]
3 years ago
14

A satellite in orbit around the Earth has a speed of 8 km/s at a given point of its orbit. If the period is 2 h, what is the alt

itude at that point?
Physics
1 answer:
zzz [600]3 years ago
7 0

Answer:

Explanation:

Using the formula;

Speed = Distance/Time

Given

Speed = 8km/s

Time = 2hr

Convert 2hr to secs

1hr = 3600secs

2hr = 2(3600)

2hr = 7200secs

Altitude is the distance

From the formula;

Distance = speed × time

Distance = 8×7200

Distance = 57200km

Hence the altitude at this point is 57,200km

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You stand on top a building 44 m tall with a water balloon. You drop the water balloon from rest. How fast is the balloon moving
Alecsey [184]
<h2>The balloon is moving when it is halfway down the building at 20.78 m/s.</h2>

Explanation:

We have equation of motion v² = u² + 2as

Initial velocity, u = 0 m/s  

Acceleration, a = 9.81 m/s²  

Displacement, s = 0.5 x 44 = 22 m

Substituting  

v² = u² + 2as

v² = 0² + 2 x 9.81 x 22

v² = 431.64

v = 20.78 m/s

Velocity at 22 m = 20.78 m/s

The balloon is moving when it is halfway down the building at 20.78 m/s.

7 0
3 years ago
How does contrasting the social structure of simple and complex societies?
Lerok [7]
Best Answer: <span>There is little or no surplus so the social inequalities are not significant and economic interaction takes place within egalitarian frame-work.

Brainliest would be awesome
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8 0
2 years ago
Brainliest please help<br><br>tell me if am right <br>if not correct me <br><br><br>​please
REY [17]

Answer:

See the answers below

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

<u>First case</u>

Vf = 6 [m/s]

Vo = 2 [m/s]

t = 2 [s]

6=2+a*2\\4=2*a\\a=2[m/s^{2} ]

<u>Second case</u>

Vf = 25 [m/s]

Vo = 5 [m/s]

a = 2 [m/s²]

25=5+2*t\\t = 10 [s]

<u>Third case</u>

Vo =4 [m/s]

a = 10 [m/s²]

t = 2 [s]

v_{f}=4+10*2\\v_{f}=24 [m/s]

<u>Fourth Case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

v_{f}=5+8*10\\v_{f}=85 [m/s]

<u>Fifth case</u>

Vf = final velocity [m/s]

Vo = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

8=v_{o}+4*2\\v_{0}=8-8\\v_{o}=0

8 0
3 years ago
What is a point value for an arrow that is on two different colors?
Whitepunk [10]
It depends on what it is closest to but I would say for instance black is 5 points and red is 6 if u land on the line 5.5
8 0
3 years ago
Space Station Suppose a space-station is designed in s shape of a torus such as the one depicted in Stanley Kubrick's "2001: A s
yaroslaw [1]

Answer:

w = 3.2 rev / min

Explanation:

For this exercise we will use the centrine acceleration equal to the acceleration of gravity

      a = v² / r

Angular and linear variables are related.

     v = w r

Let's replace

     a = w² r = g

     w = √ g / r

     r = d / 2

     r = 175/2 = 87.5 m

    w = √( 9.8 / 87.5)

    w = 0.3347 rad / s

Let's reduce to rotations per min

     w = 0.3347 rad / s (1 rov / 2pi rad) (60 s / 1 min)

     w = 3.2 rev / min

Suppose the space station rotates counterclockwise, we have two possibilities for the car

The first car turns counterclockwise (same direction of the station

     v_{c} =  w_{c} r

     [texwv_{c}[/tex] =  v_{c} / r

     [texwv_{c}[/tex] = 25.0 / 87.5

     [texwv_{c}[/tex] = 0.286 rad / s

When the two rotate in the same direction their angular speeds are subtracted

     w total = w -[texwv_{c}[/tex]

     w total = 0.3347 - 0.286

    w  total= 0.487 rad / s

The car goes in the opposite direction of the station the speeds add up

    w = 0.3347 + 0.286

    w = 0.62 rad / s

From this values ​​we can see that the person feels a variation of the acceleration of gravity, feels that he has less weight when he goes in the same direction of the season and that his weight increases when he goes in the opposite direction to the season.

3 0
3 years ago
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