Complete Question
The complete question is shown on the uploaded image
Answer:
The tension on the shank is ![T =8391.6 N](https://tex.z-dn.net/?f=T%20%3D8391.6%20N)
Explanation:
From the question we are told that
The strain on the strain on the head is ![\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m](https://tex.z-dn.net/?f=%5CDelta%20l%20%3D%200.1%20mm%2Fmm%20%3D%20%5Cfrac%7B0.1%7D%7B1000%7D%20%3D%200.1%20%2A10%5E%7B-3%7D%20m%2Fm)
The contact area is
Looking at the first diagram
At 600 MPa of stress
The strain is ![0.3mm/mm](https://tex.z-dn.net/?f=0.3mm%2Fmm)
At 450 MPa of stress
The strain is ![0.0015 mm/mm](https://tex.z-dn.net/?f=0.0015%20mm%2Fmm)
To find the stress at
we use the interpolation method
![\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma_%7B%5CDelta%20l%7D%20-%20%20%5Csigma_%7B0.0015%7D%20%7D%7B%20%5Csigma%20_%20%7B0.3%7D%20-%20%5Csigma_%7B0.0015%7D%20%7D%20%3D%20%5Cfrac%7Be_%7B%5CDelta%20l%20%7D%20%20-%20e_%7B0.0015%7D%7D%7Be_%7B0.3%7D%20-%20e_%7B%200.0015%7D%7D)
Substituting values
![\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%20_%7B%5CDelta%20l%7D%20-%20450%7D%7B600%20-%20450%7D%20%3D%20%5Cfrac%7B0.1%20-0.0015%7D%7B0.3%20-%200.0015%7D)
![\sigma _{\Delta l} -450 = 49.50](https://tex.z-dn.net/?f=%5Csigma%20_%7B%5CDelta%20l%7D%20-450%20%3D%2049.50)
![\sigma _{\Delta l} = 499.50 MPa](https://tex.z-dn.net/?f=%5Csigma%20_%7B%5CDelta%20l%7D%20%3D%20499.50%20MPa)
Generally the force on each head is mathematically represented as
![F = \sigma_{\Delta l} * A](https://tex.z-dn.net/?f=F%20%3D%20%5Csigma_%7B%5CDelta%20l%7D%20%2A%20A)
Substituting values
![F = 499.50*10^{6} * 2.8*10^{-6}](https://tex.z-dn.net/?f=F%20%3D%20499.50%2A10%5E%7B6%7D%20%2A%202.8%2A10%5E%7B-6%7D)
![=1398.6N](https://tex.z-dn.net/?f=%3D1398.6N)
Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as
![T = 6 * F](https://tex.z-dn.net/?f=T%20%3D%206%20%2A%20F)
![= 6* 1398.6](https://tex.z-dn.net/?f=%3D%206%2A%201398.6)
![T =8391.6 N](https://tex.z-dn.net/?f=T%20%3D8391.6%20N)
The temperature of the gas is 41.3 °C.
Answer:
The temperature of the gas is 41.3 °C.
Explanation:
So on combining the Boyle's and Charles law, we get the ideal law of gas that is PV=nRT. Here P is the pressure, V is the volume, n is the number of moles, R is gas constant and T is the temperature. The SI unit of pressure is atm. So we need to convert 1 Pa to 1 atm, that is 1 Pa = 9.86923×
atm. Thus, 171000 Pa = 1.6876 atm.
We know that the gas constant R = 0.0821 atmLMol–¹K-¹. Then the volume of the gas is given as 50 L and moles are given as 3.27 moles.
Then substituting all the values in ideal gas equation ,we get
1.6876×50=3.27×0.0821×T
Temperature = ![\frac{84.38}{0.268467} =314.3 K](https://tex.z-dn.net/?f=%5Cfrac%7B84.38%7D%7B0.268467%7D%20%3D314.3%20K)
So the temperature is obtained to be 314.3 K. As 0°C = 273 K,
Then 314.3 K = 314.3-273 °C=41.3 °C.
Thus, the temperature is 41.3 °C.
When a satellite is revolving into the orbit around a planet then we can say
net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have
![F_g = F_c](https://tex.z-dn.net/?f=F_g%20%3D%20F_c)
![\frac{GM_pM_s}{r^2} = \frac{M_s v^2}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BGM_pM_s%7D%7Br%5E2%7D%20%3D%20%5Cfrac%7BM_s%20v%5E2%7D%7Br%7D)
now we can say that kinetic energy of satellite is given as
![KE = \frac{1}{2}M_s v^2](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7B1%7D%7B2%7DM_s%20v%5E2)
![KE = \frac{GM_sM_p}{2r}](https://tex.z-dn.net/?f=KE%20%3D%20%5Cfrac%7BGM_sM_p%7D%7B2r%7D)
also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it
so we will have
![U = -\frac{GM_sM_p}{r}](https://tex.z-dn.net/?f=U%20%3D%20-%5Cfrac%7BGM_sM_p%7D%7Br%7D)
so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite