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Usimov [2.4K]
3 years ago
14

A meteroid is in a circular orbit 600 km above the surface of a distant planet. the planet has the same mass as Earth but has a

radius that is 90% of Earths (where Earths radius is approximately 6370 km).
The acceleration of the meteroid due to the gravitational force exerted by the planet is most nearly

A) 9 m/s^2 toward the center of the planet

B) 9 m/s^2 in the direction of the meteoroids motion

C) 10 m/s^2 toward the center of the planet

D) 10 m/s^2 in the direction of the meteoroids motion​
Physics
2 answers:
taurus [48]3 years ago
6 0
Well, the answer to this question is D) 10 m/s^2 in the direction of the meteoroids motion.
Glad I could help and have a fantastic day!
Also, can you please mark my answer as the brainliest answer?
Thanks.
Llana [10]3 years ago
3 0

Answer:

C)  10 m/s^2 toward the center of the planet

Explanation:

g = GM/r^2

The masses of earth and this planet are the same and the radius is only diffferent by 10% so as gravitational force acceleration is 9.8m/s^2 on earth, it can be estimated to be around 10m/s^2.

Using the third law, we know acceleration is toward the larger mass from attraction, thus having it point toward the center of the planet.

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nika2105 [10]
Slow moving vehicles that travel at a sped less than 25 MPH will have ...........
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7 0
3 years ago
a proton travelling along the x-axis is slowed by a uniform electric field E. at x = 20 cm, the proton has a speed of 3.5x10^6 m
AveGali [126]

Answer:

The magnitude of the electric field is 3.8 × 10⁵⁸ N/C and it is in the negative x- direction.

Explanation:

From work-kinetic energy principles, the kinetic energy change of the proton equals the work done by the electric field.

So, ΔK = W

1/2m(v₂² - v₁²) = qEd where m = mass of proton = 1.673 × 10⁻²⁷ kg, v₁ = initial speed of proton = 3.5 × 10⁶ m/s, v₂ = final speed of proton = 0 m/s, q = proton charge = + e = 1.602 × 10⁻¹⁹ C, E = electric field and d = distance moved by proton = x₂ - x₁ , x₁ = 20 cm and x₂ = 80 cm. So, d = 80 cm - 20 cm = 60 cm = 0.6 m

1/2m(v₂² - v₁²) = qEd

E = (v₂² - v₁²)/2mqd

substituting the values of the variables into E, we have

E = ((0 m/s)² - (3.5 × 10⁶ m/s)²)/(2 × 1.673 × 10⁻²⁷ kg × 1.602 × 10⁻¹⁹ C × 0.6 m)

E = - 12.25 × 10¹² m²/s² ÷ 3.22 × 10⁻⁴⁶

E = -3.8 × 10⁵⁸ N/C

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6 0
3 years ago
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slope of bird1 is 2.8/2.7
slope of bird2 is 2.7/2/8
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As the distance between two charged objects increases, the strength of the electrical force between the objects
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