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yuradex [85]
4 years ago
7

As a student in a physics lab, you are asked to classify two springs. Right away you notice some differences. The first spring,

spring 1, is very easy to compress and pliable. The second spring, spring 2, is stiffer and more difficult to deform. What qualitative statement can you make about their spring constants?
A. No inference can be made, the student needs to take a measurement.
B. The spring constant of spring 1 is larger than spring 2.
C. The spring constant of spring 2 is larger than spring 1.
D. The spring constants are the same because spring 1 has more displacement, while spring 2 requires more force.
Physics
1 answer:
strojnjashka [21]4 years ago
3 0

Answer:

C. The spring constant of spring 2 is larger than spring 1.

Explanation:

As we know that spring constant also called the stiffness constant of a spring is the value of force required for the deformation of the spring by a unit length.

Mathematically given as:

k=\frac{F}{\delta x}

where:

k= stiffness constant

F= force on the spring

\delta x= change in length of the spring

It is given in the question that spring 1 is easily compressible and requires less force to compress as compared to spring 2 and hence it has lesser stiffness constant than that of spring 2.

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A solid, horizontal cylinder of mass 10.6 kg and radius 1.00 m rotates with an angular speed of 8.00 rad/s about a fixed vertica
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Answer:

The final angular speed is 7.71 rad/s

Explanation:

Given

Cylinder mass, M = 10.6 kg

Cylinder radius, R = 1.00 m

Angular speed, w = 8.00 rad/s.

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Radius, r = 0.900 m

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While the moment of inertia if the putty is mr².

Initial Momentum of the system = Initial momentum of the cylinder =

Li = Iw --- Substitute ½MR² for I

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By

Substituton

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Li = 42.4kgm²/s

Calculating the final momentum of the system.

First we calculate the final momentum of the cylinder

Li = Iw --- Substitute ½MR² for I

Li = ½MR²wf where wf = final angular speed

By

Substituton

Li = ½ * 10.6 * 1² wf

Li = 5.3w kgm²/s

Then we calculate the final momentum of the putty

Final Momentum of the putty =

L2 = Iwf --- Substitute mr² for I;

L2 = mr²wf --- By Substituton

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Final momentum = Li + L2

Lf = (5.3wf + 0.2025wf) kgm²/s

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By conservation of momentum

Li = Lf

Where Li = 42.4kgm²/s and Lf = 5.5025wf kgm²/s

So, we have

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wf = 42.4/5.5025

wf = 7.71 rad/s

Hence, the final angular speed is 7.71 rad/s

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