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3 years ago

Which object moves in simple harmonic motion?\

Physics

1 answer:

Rina8888 [55]3 years ago

3 0

<h2>Answer:Stretched rubber band with an mass</h2>

Explanation:

Simple harmonic motion requires a restoring force.

Simple harmonic motion should be periodic.

Option A:

When a rubber band is stretched,the internal forces of the rubber band pulls the band inside.So,internal forces are the restoring forces in a rubber band.

The motion is periodic as well since the the rubber band makes to an fro motion expanding ans contracting.

So,this performs SHM.

Option B:

There is no restoring force and no periodicity.

So,this is not SHM.

Option C:

There is no restoring force and no periodicity.

So,this is not SHM.

Option D:

There is periodicity but no restoring force.

So,this is not SHM.

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2 years ago

If the distance between two masses is tripled, the gravitational force between changes by a factor of:_______

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Answer:

option A

Explanation:

given,

distance between two masses is doubled

new distance, r' = 3 r

using gravitational force equation

$F = \dfrac{GMm}{r^2}$ ............(1)

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now from the given condition

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$F' = \dfrac{GMm}{9r^2}$

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3 years ago

Object A is 71 degrees and object B is 75 degrees how will thermal energy flow

Tasya [4]

Given :

Object A is 71 degrees and object B is 75 degrees .

To Find :

How will thermal energy flow.

Solution :

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2 years ago

An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

1.228 x $10^{-6}$ mm

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

==> $\frac{180*10^{3}}{85*10^{9}} *( \frac{500}{1256.8} + \frac{100}{549.85})$ = 1.228 x $10^{-6}$ mm

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