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3 years ago

Difference between rest and motion

1 answer:

3 0

Rest - it is the state in which body doesn’t move from it’s place

motion - it is the state in which body moves from it’s place

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PLEASE HELP!!!!! Will give brainliest to best answer

PolarNik [594]

Answer:

the answer is d

Explanation:

I took the test

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4 years ago

A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was

Rudiy27

Answer:

$X=0.0389m$

Explanation:

From the question we are told that:

Period of spring $T_s=2.25s$

Initial Position of Mass $x=0.0480m$

Final Mass period $T_f=5.85s$

Generally the equation for the Mass location is mathematically given by

$X=xcos*\frac{2\pi T_s}{T_f}$

$X=0.048*cos*\frac{2\pi 5.85}{2.25}$

$X=0.0389m$

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3 years ago

A thin rod of length 1.4 m and mass 180 g is suspended freely from one end. It is pulled to one side and then allowed to swing l

prisoha [69]

Answer:

$K.E = 0.1905 J$

Explanation:

From the question we are told that:

Length $L=1.4m$

Mass $m=180g$

Angular Velocity $\omega=1.80rads/s$

Generally the equation for Kinetic energy K.E is mathematically given by

$K.E =0.5 (1/3 ML^2 )w^2$

$K.E =0.5 ( 1/3 * 0.18 * 1.4^2 ) 1.8^2$

$K.E = 0.1905 J$

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3 years ago

PLEASE HELP ME !!!!!

telo118 [61]

Answer: radiant energy

Explanation:

The energy in electromagnetic waves is sometimes called radiant energy.

4 0

3 years ago

A 0.560 kg snowball is fired from a cliff 14.2 m high with an initial velocity of 13.3 m/s, directed 26.0° above the horizontal.

enot [183]

Answer:

a) v = 21.34 m/s

b) v = 21.34 m/s

c) v = 21.34 m/s

Explanation:

Mass of the snowball, m = 0.560 kg

Height of the cliff, h = 14.2 m

Initial velocity of the ball, u = 13.3 m/s

θ = 26°

The speed of the slow ball as it reaches the ground, v = ?

The initial Kinetic energy of the snow ball, $KE_{0} = 0.5 mu^{2}$

Potential energy of the snow ball at the given height, PE = mgh

Final Kinetic energy of the ball as it reaches the ground, $KE_{f} = 0.5mv^{2}$

a) Using the principle of energy conservation,

$KE_{0} + PE = KE_{f} \\0.5mu^{2} + mgh = 0.5mv^{2}\\v^{2} =2( 0.5u^{2} + gh)\\v^{2} =u^{2} + 2gh\\v = \sqrt{u^{2} + 2gh} \\v = \sqrt{13.3^{2} + 2*9.8*14.2}\\v = 21.34 m/s$

b) The speed remains v = 21.34 m/s since it is not a function of the angle of launch

c)The principle of energy conservation used cancels out the mass of the object, therefore the speed is not dependent on mass

v = 21. 34 m/s

7 0

4 years ago

Difference between rest and motion​

Difference between rest and motion