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PilotLPTM [1.2K]
3 years ago
14

What happens when you push a spring? How is this different than pulling it? (Hooke’s Law)

Physics
2 answers:
Yuki888 [10]3 years ago
7 0

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.

Furkat [3]3 years ago
4 0

When you push a spring you are putting force on it, or rather forcing the spring onto its self. This is different than pulling the spring because when you push it and let go it will spring but if you pull it its would not move as far as it would if you push it. Pulling the spring would most likely deform the spring will pushing it would cause it to spring back into place.

<u>Hooke's law states that the force that extends or compresses a spring by some distance scales linearly with respect to that distance.</u>

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Suppose you take a 50gram ice cube from the freezer at an initial temperature of -20°C. How much energy would it take to complet
notsponge [240]

Answer:

The amount of energy required is 152.68\times 10^{3}Joules

Explanation:

The energy required to convert the ice to steam is the sum of:

1) Energy required to raise the temperature of the ice from -20 to 0 degree Celsius.

2) Latent heat required to convert the ice into water.

3) Energy required to raise the temperature of water from 0 degrees to 100 degrees

4) Latent heat required to convert the water at 100 degrees to steam.

The amount of energy required in each process is as under

1) Q_1=mass\times S.heat_{ice}\times \Delta T\\\\Q_1=50\times 2.05\times 20=2050Joules

where

'S.heat_{ice}' is specific heat of ice =2.05J/^{o}C\cdot gm

2) Amount of heat required in phase 2 equals

Q_2=L.heat\times mass\\\\\therefore Q_{2}=334\times 50=16700Joules

3) The amount of heat required to raise the temperature of water from 0 to 100 degrees centigrade equals

Q_3=mass\times S.heat\times \Delta T\\\\Q_1=50\times 4.186\times 100=20930Joules

where

'S.heat_{water}' is specific heat of water=4.186J/^{o}C\cdot gm

4) Amount of heat required in phase 4 equals

Q_4=L.heat\times mass\\\\\therefore Q_{4}=2260\times 50=113000Joules\\\\\\\\\\\\Thus the total heat required equals Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\\\Q=152.68\times 10^{3}Joules

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