Answer:
Part a)
![a= 0.32 m/s^2](https://tex.z-dn.net/?f=a%3D%200.32%20m%2Fs%5E2)
Part b)
![F_c = 3.6 N](https://tex.z-dn.net/?f=F_c%20%3D%203.6%20N)
Part c)
![F_c = 5.5 N](https://tex.z-dn.net/?f=F_c%20%3D%205.5%20N)
Explanation:
Part a)
As we know that the friction force on two boxes is given as
![F_f = \mu m_a g + \mu m_b g](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20m_a%20g%20%2B%20%5Cmu%20m_b%20g)
![F_f = 0.02(10.6 + 7)9.81](https://tex.z-dn.net/?f=F_f%20%3D%200.02%2810.6%20%2B%207%299.81)
![F_f = 3.45 N](https://tex.z-dn.net/?f=F_f%20%3D%203.45%20N)
Now we know by Newton's II law
![F_{net} = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma)
so we have
![F_p - F_f = (m_a + m_b) a](https://tex.z-dn.net/?f=F_p%20-%20F_f%20%3D%20%28m_a%20%2B%20m_b%29%20a)
![9.1 - 3.45 = (10.6 + 7) a](https://tex.z-dn.net/?f=9.1%20-%203.45%20%3D%20%2810.6%20%2B%207%29%20a)
![a = \frac{5.65}{17.6}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B5.65%7D%7B17.6%7D)
![a= 0.32 m/s^2](https://tex.z-dn.net/?f=a%3D%200.32%20m%2Fs%5E2)
Part b)
For block B we know that net force on it will push it forward with same acceleration so we have
![F_c - F_f = m_b a](https://tex.z-dn.net/?f=F_c%20-%20F_f%20%3D%20m_b%20a)
![F_c = \mu m_b g + m_b a](https://tex.z-dn.net/?f=F_c%20%3D%20%5Cmu%20m_b%20g%20%2B%20m_b%20a)
![F_c = 0.02(7)(9.8) + 7(0.32)](https://tex.z-dn.net/?f=F_c%20%3D%200.02%287%29%289.8%29%20%2B%207%280.32%29)
![F_c = 3.6 N](https://tex.z-dn.net/?f=F_c%20%3D%203.6%20N)
Part c)
If Alex push from other side then also the acceleration will be same
So for box B we can say that Net force is given as
![F_p - F_f - F_c = m_b a](https://tex.z-dn.net/?f=F_p%20-%20F_f%20-%20F_c%20%3D%20m_b%20a)
![9.1 - 0.02(7)(9.8) - F_c = 7(0.32)](https://tex.z-dn.net/?f=9.1%20-%200.02%287%29%289.8%29%20-%20F_c%20%3D%207%280.32%29)
![F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)](https://tex.z-dn.net/?f=F_c%20%3D%209.1%20-%200.02%20%287%29%289.8%29%20-%207%280.32%29)
![F_c = 5.5 N](https://tex.z-dn.net/?f=F_c%20%3D%205.5%20N)
Answer:
to turn red litmus blue and react with acid to form salt
I believe your answer is d).
Hope you find this helpful! :)
Answer:
![P=58.3kW](https://tex.z-dn.net/?f=P%3D58.3kW)
Explanation:
Given data
Length L=2.5 m
Radius R=d/2=30/2 = 15 mm
Torque based on allowable stress
Allowable shear stress τ=50 Mpa
Allowable torque T=(π/2)τc³
![T=\frac{\pi }{2}(50*10^{6} )(0.015)^{3} \\ T=264.94N.m](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B%5Cpi%20%7D%7B2%7D%2850%2A10%5E%7B6%7D%20%29%280.015%29%5E%7B3%7D%20%5C%5C%20T%3D264.94N.m)
Torque based on allowable angle of twist
Allowable Angle of twist
Ф=7.5°
Ф=7.5×(π/180)=130.90×10⁻³ rad
Allowable torque
T=(GJФ)/L
T=(G(π/2)c⁴)Ф)/L
T=(πGc⁴Ф)/2
![T=\frac{\pi (77.2*10^{9} )(0.015)^{4}(130.90*10^{-3} ) }{2}\\ T=265.07N.m](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B%5Cpi%20%2877.2%2A10%5E%7B9%7D%20%29%280.015%29%5E%7B4%7D%28130.90%2A10%5E%7B-3%7D%20%29%20%7D%7B2%7D%5C%5C%20T%3D265.07N.m)
Maximum Power Transmitted
Maximum power transmitted is given by
![P=2\pi fT\\P=2\pi (35)(265.07)\\P=58262.386watt\\P=58.3kW](https://tex.z-dn.net/?f=P%3D2%5Cpi%20fT%5C%5CP%3D2%5Cpi%20%2835%29%28265.07%29%5C%5CP%3D58262.386watt%5C%5CP%3D58.3kW)