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xenn [34]
2 years ago
15

The amplitude of a mechanical wave shows What?

Physics
1 answer:
Anuta_ua [19.1K]2 years ago
5 0
I’m not sure what you are asking
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A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = 6.38\times 10^6\ m

The gravitational force will balance the centripetal force

\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}

T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}

From Kepler's law we have relation

T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m

Distance from the center of the Earth would be

42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}

8 0
3 years ago
Can someone help? Please?
zubka84 [21]

Answer:

A. Speed

Explanation:

Speed is the magnitude of velocity, which is given in the question. Velocity is a vector quantity and therefore has both a magnitude and a direction. Only the former is implied in the question.

5 0
2 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
A6 kg mass moving at 10m/s collides with a 4 kg mass moving in the
Nat2105 [25]

Answer:

Explanation:

mdeemmkdkwdmwdmw

4 0
3 years ago
A BODY STARTING FROM REST MOVES WITH CONSTANT ACCELERATON. what is the ratio of distance covered by the body during the fifth se
Art [367]

Answer:

9/25

Explanation:

Distance covered in the first 5 seconds:

Δx = v₀ t + ½ at²

Δx₀₋₅ = (0) (5) + ½ a (5)²

Δx₀₋₅ = 25a/2

Distance covered in the first 4 seconds:

Δx = v₀ t + ½ at²

Δx₀₋₄ = (0) (4) + ½ a (4)²

Δx₀₋₄ = 8a

So the distance covered during the 5th second is:

Δx₅ = 25a/2 − 8a

Δx₅ = 9a/2

So the ratio of the distance covered during the 5th second to the distance covered in the first 5 seconds is:

Δx₅ / Δx₀₋₅

(9a/2) / (25a/2)

9/25

8 0
3 years ago
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