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adoni [48]
3 years ago
13

A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,

how much elastic potential energy is stored in the spring? 700 J 1050 J 1575 J 4375 J
Physics
2 answers:
Vlad [161]3 years ago
6 0

Answer:

1575 J

Explanation:

The elastic potential energy of a string is given by:

E=\frac{1}{2}k \Delta x^2

where

k is the spring constant

\Delta x is the stretching/compression of the string with respect to its natural length

For the spring in the problem, we have:

- Spring constant: k=1400 N/m

- Stretching: \Delta x=2.5 m -1.0 m=1.5 m

Therefore, the elastic potential energy stored in the spring is

E=\frac{1}{2}(1400 N/m)(1.5 m)^2=1575 J

Elena L [17]3 years ago
5 0

The answer is C. 1575 J. I just took this review.

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irinina [24]
PV=nRT
(P)(.010)=(n)(.08201)(0)
(v1/t1)=(v2/t2)
(.010/t1)=(v2/0)
The volume would be zero 
7 0
3 years ago
When sound is created it travels to the car through a
Rasek [7]

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This could maybe help you with your answer.

4 0
3 years ago
A ball is traveling 24° above the horizontal at a speed of 12 m/s. What is the vertical component of its speed?
victus00 [196]

Answer:

4.88 m/s

Explanation:

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Horizontal is   12 * cos 24

3 0
2 years ago
d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s
Harrizon [31]

Answer:

a' = 0.35 m/s^2

Explanation:

Let say the empty wagon has mass "M"

now by newton's II Law we will have

F = Ma

now it is given that empty wagon is pulled with acceleration 1.4 m/s/s

now we will have

F = 1.4 M

now a child of mass three times the mass of wagon is sitting on the empty wagon

so here we have

F = (M + 3M) a

1.4 M = 4M a'

so we have

a' = 0.35 m/s^2

8 0
3 years ago
A 55.0-g sample of hot metal initially at 99.5oC was added to 40.0 g of water in a Styrofoam coffee cup calorimeter. The water a
Kaylis [27]

Answer:

Cp= 0.44 J/g.C

This is heat capacity of metal.

Explanation:

From energy conservation

Heat lost by metal = Heat gain by water +Heat gain by  calorimeter

Because here temperature of metal is high that is why it loose the heat.The temperature of water and  calorimeter is low that is why they gain the heat.

final temperature is T= 30.5 C

We know that sensible heat transfer given as

Q= m Cp ΔT

m=Mass

Cp=Specific heat capacity

ΔT=Temperature difference

By putting the values

55 x Cp ( 99.5 - 30.5) = 40 x 4.184 ( 30.5- 21 ) + 10 x ( 30.5 - 21)

Cp ( 99 .5- 30.5) = 30.65

Cp= 0.44 J/g.C

This is heat capacity of metal.

4 0
3 years ago
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