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Svetllana [295]
3 years ago
7

The planet Krypton has a mass of 7.6 × 1023 kg and radius of 1.7 × 106 m. What is the acceleration of an object in free fall nea

r the surface of Krypton? The gravitational constant is 6.6726 × 10−11 N · m2 /kg2 . Answer in units of m/s 2 .
Physics
1 answer:
olga55 [171]3 years ago
5 0

Answer:

17.55 m/s²

Explanation:

Parameters given:

Mass of Krypton, M = 7.6 * 10^23 kg

Radius, R = 1.7 * 10^6 m

Gravitational constant, G = 6.6726 * 10^(-11) Nm²/kg²

Acceleration due to gravity of planet of mass M is given as:

g = GM/R²

Since the object is close to the surface of Krypton, we can say that the distance from the Centre of Krypton is the radius of the planet Krypton.

Therefore,

g = (6.6726 * 10^(-11) * 7.6 * 10^23)/(1.7 * 10^6)²

g = 17.55 m/s²

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Answer:

See Explanation Below

Explanation:

This question is incomplete.

I'll answer this question on general terms. You'll get your result if you apply the steps I'll highlight below.

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The simple translation of this law is that, the sum of current in the close loop gives us the desired result.

Rephrasing your question;

Three currents, (I1 = +3A, I2 = +4A and I3 = -5A) are passing through a surface bounded by a closed path. The currents have different values and directions. According to Ampere’s law, what is the value of I on the right side of this equation?

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