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4 years ago

7

What is the ratio of magnitudes of their angular velocities, ω1/ω2? express your answer in terms of the variables r1 and r2?

1 answer:

kicyunya [14]4 years ago

8 0

<span>We are using the formula: v = rĎ‰ Please consider that they have the same tangential velocity at their edges. Therefore we can equate: r1 Ď‰1 = r2 Ď‰2 And re-arrange: Ď‰1 / Ď‰2 = r2 / r1</span>

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If the distance between the two mass double what happens to the gravitational force

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If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces.

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3 years ago

The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.

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2 years ago

The elements selenium and oxygen are both poor electrical conductors. In which section of the periodic table are they

murzikaleks [220]

Answer:

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Explanation:

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3 years ago

What is the mechanical advantage of a lever that has an input arm of 6 meters and an output arm of 2 meters

Setler79 [48]

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3 years ago

You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th

PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

$Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}$

where,

Y = Young's Modulus

F = force exerted by the weight = $m\times g$

m = mass of the ball = 10 kg

g = acceleration due to gravity = $9.81m/s^2$

l = length of wire = 2.6 m

A = area of cross section = $\pi r^2$

r = radius of the wire = $\frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m$ (Conversion factor: 1 m = 1000 mm)

$\Delta l$ = change in length = 1.99 mm = $1.99\times 10^{-3}m$

Putting values in above equation, we get:

$Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2$

Hence, the Young's modulus for the wire is $6.378\times 10^{10}N/m^2$

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4 years ago