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nadya68 [22]
2 years ago
8

A satellite, moving in an elliptical orbit, is 368 km above Earth's surface at its farthest point and 164 km above at its closes

t point. (a) Calculate the semimajor axis of the orbit. Incorrect: Your answer is incorrect. m (b) Calculate the eccentricity of the orbit. Incorrect: Your answer is incorrect. Did you find the semimajor axis a from the greatest and smallest radii
Physics
1 answer:
Marat540 [252]2 years ago
3 0

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

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The acceleration of the wagon along the ground is 3.6 m/s².

To solve the problem above, we need to use the formula of acceleration as related to force and mass.

Acceleration: This can be defined as the rate of change of velocity.

⇒ Formula:

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⇒ Where:

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⇒ make a the subject of equation 1

  • a = Fcos∅/m..................... Equation 2

From the question,

⇒ Given:

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⇒ Substitute these values into equation 2

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Hence, The acceleration of the wagon along the ground is 3.6 m/s²

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What are two reasons why density is a useful physical property for identifying substances?
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An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.
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Answer:

Value of electric field along the axis and equitorial axis  E=31.25\ N/c and E = 15.625\ N/c respectively.

Explanation:

Given :

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Magnitude of charges , q=1\ nC = 10^{-9}\ C.

Dipole moment , p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

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Putting all values and r=12\times 10^{-2}\ m.

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Case B) (x,y) = (0 cm, 12.0 cm) :

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Putting all values and r=12\times 10^{-2}\ m.

We get , E = 15.625\ N/c.

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