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ohaa [14]
3 years ago
6

A(n) 96.1 g ball is dropped from a height of 59.1 cm above a spring of negligible mass.The ball compresses the spring to a maxim

um displacement of 4.75403 cm. The acceleration of gravity is 9.8 m/s^2. Calculate the spring force constant k. Answer in units of N/
Physics
1 answer:
Serhud [2]3 years ago
4 0

Answer:

Explanation:

Mass of ball Is m=96.1g=0.0961kg

Height above spring is 59.1cm

L=0.591m

Extension of the spring is 4.75403cm

e=0.0475403m

Then the distance the ball traveled is H=L+e

H=0.591+0.0475403

H=0.6385403m

Then, the potential energy of the ball is given as

P.E=mgh

P.E=0.0961×9.81×0.6385403

P.E=0.602J

From conservation of energy, energy cannot be created nor destroy but can be transferred from one form to another

Then, the P.E is transferred to the work done by the spring

Then, Work done by spring is given as

W=½ke²

W=P.E=½×k×0.0475403²

0.602=½×k×0.0475403²

k=0.602×2/0.0475403²

k=532.72N/m

The spring constant is 532.72 N/m

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<u>Answer:</u> The angle of diffraction is 0.498°

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3 years ago
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
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Answer:

The object will travel 675 m during that time.

Explanation:

A body moves with constant acceleration motion or uniformly accelerated rectilinear motion (u.a.r.m) when the path is a straight line, but the velocity is not necessarily constant because there is an acceleration.

In other words, a body performs a u.a.r.m when its path is a straight line and its acceleration is constant. This implies that the speed increases or decreases uniformly.

In this case, the position is calculated using the expression:

x = xo + vo*t + ½*a*t²

where:

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  • v0 is the initial velocity.
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  • t is the time interval in which the motion is studied.

In this case:

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Replacing:

x= 0 + 0*15 s + ½*6 \frac{m}{s^{2} }*(15s)²

Solving:

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x= 675 m

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