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malfutka [58]
3 years ago
14

10) T F Two unequal weights are connected by a massless string which passes over a frictionless pulley. If the pulley has no app

reciable mass, the tension in the string is the same on both sides of the pulley; but if the pulley has mass, the tension will not be the same on both sides of the pulley.
Physics
2 answers:
neonofarm [45]3 years ago
3 0

Answer:

2

Explanation:

2

swat323 years ago
3 0

Let tension developed in the string be T.

Equations of motion are as follows-

T=M2a

M1g−T=M1a

which gives, a=gM1+M2M1

<h2>for more</h2>

brainly.com/question/12497530

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a) B. The force will be nine times larger than the initial value.

b) D. 3W/2

Explanation:

a) If we can take both masses as point masses, the attractive force between them, is given by Newton's Universal Law of gravitation:

Fg = G*m₁*m₂ / r₁₂² (1)

where r₁₂ is the distance between the center of the masses.

If we reduce the distance in such a way that r₁₂f = r₁₂₀/3, replacing in (1), we get:

Fgf = G*m₁*m₂  / (r₁₂₀/3)² = 9* (G*m₁*m₂/ r₁₂₀²) = 9* Fg₀

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b) We call weight, to the force that earth exerts on any mass on the planet, based on the application of Newton's 2nd Law to the gravitational force between Earth and the mass, as follows:

Fg = G*m*me / re² = m*a

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If we apply the same considerations to the same mass on the surface of a planet with a mass equal to one-sixth the mass of Earth, and a radius one third that of Earth, we can apply the same equation as above:

Fgp = G*m*(me/6) / (re/3)² = m*ap

⇒ap = gp = (G*me/re²)*(9/6) = 3/2* ge

As the mass is the same, we conclude that the gravitational force exerted by the unknown planet on the mass (which we call weight) is 3/2 times the one experimented on Earth's  surface.

So, the right answer is D. 3/2W.

8 0
3 years ago
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