Answer:
Work done, W = 0.0219 J
Explanation:
Given that,
Force constant of the spring, k = 290 N/m
Compression in the spring, x = 12.3 mm = 0.0123 m
We need to find the work done to compress a spring. The work done in this way is given by :
![W=\dfrac{1}{2}kx^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1%7D%7B2%7Dkx%5E2)
![W=\dfrac{1}{2}\times 290\times (0.0123)^2](https://tex.z-dn.net/?f=W%3D%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%20290%5Ctimes%20%280.0123%29%5E2)
W = 0.0219 J
So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.
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Answer:
Element Chart
Explanation:
It is a chart that provides many different elements.
Answer : 6.022• 10^23 atoms of potassium
Answer:
Ae/A* = 1.115
Explanation:
Let the reservoir pressure be ![p_0](https://tex.z-dn.net/?f=p_0)
Let the exit pressure be ![p_e](https://tex.z-dn.net/?f=p_e)
Ratio of reservoir pressure and exit pressure
![\frac{p_o}{p_e} = \frac{1}{0.3143}](https://tex.z-dn.net/?f=%5Cfrac%7Bp_o%7D%7Bp_e%7D%20%3D%20%5Cfrac%7B1%7D%7B0.3143%7D)
= 3.182
For the above value of pressure ratio
Obtain the area ratio from the isentropic flow table
Ae/A* = 1.115
The value of pressure ratio is Ae/A* = 1.115