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3 years ago

If a 2 kg spring is compressed by exerting a force of 10 newtons a distance of 50 cm how much work is done?

1 answer:

5 0

Convert cm into meters,
50/100=0.5m

work done= 10x0.5
=5J

we dont consider the weight of the spring as it acts downwards.

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A golf ball with an initial angle of 34° lands exactly 240 m down the range on a level course.

g100num [7]

Answer:50.39 m/s,

40.46 m

Explanation:

Given

launch angle $=34^{\circ}$

Range=240 m

We know that Range $=\frac{u^2sin2\theta }{g}$

$240=\frac{u^2sin(68)}{9.81}$

u=50.39 m/s

(b)maximum height of projectile is given by

$H=\frac{u^2sin^2\theta }{2g}$

$H=\frac{50.39^2(sin34)^2}{2\times 9.81}$

H=40.46 m

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mamaluj [8]

The energy that generates wind on an individual basis originates with
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3 years ago

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monitta

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2 years ago

Trevor places a mass on a spring. When he releases it, the mass and spring move in simple harmonic motion.

777dan777 [17]

The correct answer is D He could pull the mass down farther.

Amplitude, is the maximum displacement or distance moved by a point on a vibrating body or wave measured from its equilibrium position. So to increase the amplitude the spring should be pulled down further which can increase the amplitude.

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3 years ago

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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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3 years ago