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svlad2 [7]
3 years ago
9

Two objects of the same mass are on two different planets. Planet A has a force of gravity that is weaker than that of planet B.

How will the weights of the objects compare to each other?
The weights of the objects will be the same.
The weight of the object on planet A will be greater than the weight of the object on planet B.
The weight of the object on planet A will be less than the weight of the object on planet B.
This cannot be determined without knowing the actual mass of the objects.
Physics
1 answer:
iris [78.8K]3 years ago
7 0
C) The weight of the object on planet A will be less than the weight of the object on planet B.


This happened because: when gravitational pull of a planet is weak, an object will weigh less as a result of less gravity forcing it down to the ground.
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Answer:

Planetesimals  (Ex, Mercury, Venus , Mars & Earth)

Explanation:

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Elaborate on the reason(s) that matter is said to move even as in a solid state.
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<u>Answer:</u>

The matter does not move in solid state but vibrates.

<u>Explanation:</u>

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Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

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Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

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b.only when the current in the first coil changes.

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