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Radda [10]
1 year ago
8

A 0.20 kg mass attached to the end of a spring causes it to stretch 3.0 cm. What is the spring constant? What is the potential e

nergy of the spring?
Physics
1 answer:
SVEN [57.7K]1 year ago
8 0

Given that the mass is m = 0.2 kg and the displacement is x = 3 cm = 0.03 m

We have to find the spring constant and potential energy.

The spring constant can be calculated by the formula

\begin{gathered} F=\text{ kx} \\ mg\text{ = kx} \\ k\text{ = }\frac{mg}{x} \end{gathered}

Here, k is the spring constant.

g = 9.8 m/s^2 is the acceleration due to gravity.

Substituting the values, the spring constant will be

\begin{gathered} k=\frac{0.2\times9.8}{0.03} \\ =\text{ 65.33 N/m } \end{gathered}

The potential energy can be calculated as

\begin{gathered} U=\frac{1}{2}kx^2 \\ =\frac{1}{2}\times65.33\text{ }\times(0.03)^2 \\ =\text{ 0.0293 J} \end{gathered}

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A 12,000kg. railroad car is traveling at +2m/s when it
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<u>Answer:</u>

The final velocity of the two  railroad cars is 1.09 m/s

<u>Explanation:</u>

Since we are given that the two cars lock together it shows that the collision is inelastic in nature. The final velocity due to inelastic collision is given by  

\mathrm{V}=\frac{V 1 M 1+V 2 M 2}{M 1+M 2}

where

V= Final velocity

M1= mass of the first object in kgs = 12000

M2= mas of the second object in kgs = 10000

V1= initial velocity of the first object in m/s = 2m/s

V2= initial velocity of the second object in m/s = 0 (given at rest)

Substituting the given values in the formula we get

V = 2×12000 + 0x100012000 + 10000= 2400022000= 1.09 m/s  

\mathrm{V}=\frac{2 \times 1200+0 \times 1000}{12000+10000}=\frac{24000}{22000}=1.09 \mathrm{m} / \mathrm{s}

Which is the final velocity of the two  railroad cars

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3 years ago
Which statement best explains the path light takes as it travels? A. Light takes a curved path through matter and takes a straig
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State the name of the process in each of the following changes of state:
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Answer:

A. Melting

B. Freezing

C. Condensing

D. Boiling

Explanation:

4 0
3 years ago
A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t
Evgen [1.6K]

Answers:

a) 65.075 kgm/s

b) 10.526 s

c) 61.82 N

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

I=\Delta p=p_{2}-p_{1} (1)

Where:

I is the impulse

\Delta p is the change in momentum

p_{2}=mV_{2} is the final momentum of the ball with mass m=0.685 kg and final velocity (to the right) V_{2}=57 m/s

p_{1}=mV_{1} is the initial momentum of the ball with initial velocity (to the left) V_{1}=-38 m/s

So:

I=\Delta p=mV_{2}-mV_{1} (2)

I=\Delta p=m(V_{2}-V_{1}) (3)

I=\Delta p=0.685 kg (57 m/s-(-38 m/s)) (4)

I=\Delta p=65.075 kg m/s (5)

<h3>b) Time </h3>

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 1.0 cm=0.01 m:

V_{2}=V_{1}+at (6)

V_{2}^{2}=V_{1}^{2}+2ad (7)

Where:

a is the acceleration

d=0.01 m is the length the ball was compressed

t is the time

Finding a from (7):

a=\frac{V_{2}^{2}-V_{1}^{2}}{2d} (8)

a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)} (9)

a=90.25 m/s^{2} (10)

Substituting (10) in (6):

57 m/s=-38 m/s+(90.25 m/s^{2})t (11)

Finding t:

t=1.052 s (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force F is proportional to the variation of momentum  \Delta p in time  \Delta t:

F=\frac{\Delta p}{\Delta t} (13)

F=\frac{65.075 kgm/s}{1.052 s} (14)

Finally:

F=61.82 N

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