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andrew-mc [135]
2 years ago
5

If a car driving down the road doubles its speed what happens to its kinetic energy?

Physics
1 answer:
ivanzaharov [21]2 years ago
7 0

Answer:

Kinetic energy doubles

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Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .
svet-max [94.6K]

The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

<h3>Speed of the satellite</h3>

v = √GM/r

where;

  • M is mass of Earth
  • G is universal gravitation constant
  • r is distance from center of Earth = Radius of earth + 4930 km

v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]

v = 5,916.36 m/s

Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

Learn more about speed here: brainly.com/question/6504879

#SPJ1

5 0
2 years ago
Anyone know how to do this?
Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

If it's wrong I'm truly sorry.

6 0
3 years ago
Tectonic plates are large segments of the earth's crust that move slowly. Suppose one such plate has an average speed of 6.0 cm
faltersainse [42]

Answer:

1.35×10⁻⁷ m

37.278 mi/My

Explanation:

Speed of the tectonic plate= 6 cm/yr

Converting to seconds

6=\frac{6}{365.25\times 24\times 60\times 60}

So in one second it will move

\frac{6}{365.25\times 24\times 60\times 60}

In 71 seconds

71\times \frac{6}{365.25\times 24\times 60\times 60}=1.35\times 10^{-5}\ cm

The tectonic plate will move 1.35×10⁻⁵ cm or 1.35×10⁻⁷ m

Convert to mi/My

1 cm = 6.213×10⁻⁶ mi

1 M = 10⁶ years

6\times 6.213\times 10^{-6}\times 10^6=37.278\ mi/My

Speed of the tectonic plate is 37.278 mi/My

7 0
3 years ago
Define investigation to show its scientific meaning.
Murrr4er [49]

Answer:

the action of investigating something or someone; formal or systematic examination or research.

Explanation:

This definition is provided by Oxford Languages

7 0
3 years ago
A stone is thrown vertically upwards with a speed of 30.0 m/s.
matrenka [14]

a)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = final position of stone = 20.0 meters

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = ?

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

v² = 30² + 2 (- 9.8) (20 - 0)

v = 22.5 m/s


b)

Y₀ = initial position of the stone at the time of launch = 0 m

Y = maximum height gained

a = acceleration = - 9.8 m/s²

v₀ = initial speed of stone at the time of launch = 30.0 m/s

v = final speed = 0 m/s

Using the equation

v² = v₀² + 2 a (Y - Y₀)

inserting the values

0² = 30² + 2 (- 9.8) (Y - 0)

Y = 46 m



6 0
3 years ago
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