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Vedmedyk [2.9K]

3 years ago

9

A spring has a spring constant of 80 N/m. How much energy is stored in the spring when it is compressed 0.2 m past its natural l

ength?

1 answer:

Paha777 [63]3 years ago

5 0

The energy of a compressed (or stretched) spring is given by

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

This gives

$E= \frac{1}{2}80* 0.2^{2} =1.6J$

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Read 2 more answers

If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2

kaheart [24]

Answer:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

$Q = -kA \frac{\Delta T}{\Delta x}$ (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

$\Delta T$ is the temperature of difference that we want to find

$\Delta x=2.5 cm =0.025 m$ represent the thickness of the material

If we solve $\Delta T$ in absolute value from the equation (1) we got:

$\Delta T =\frac{Q \Delta x}{Ak}$

First we convert 3KW to W and we got:

$Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W$

And we have everything to replace and we got:

$\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K$

So then the difference of temperature across the material would be $\Delta T = 375 K$

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3 years ago