Remote?? maybe I’m not really sure
Answer:
2.5 is the required details
Answer:
The diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm
Answer:
Tire rotation is the least likely cause of tire wear. So, the option D is correct.
Explanation:
Step1
Under-inflation is the process of tire failure under low pressure. This contributes the wear on tire.
Step2
On breaking, kinetic energy changes to heat energy because of rubbing of tire. So, rubbing action increases the wear on the tire.
Step3
Acceleration on the vehicle increases the rubbing action as well as the wear and tear on the tire. So, acceleration is an also a major cause of tire wear.
Step4
Tire rotation has least amount of wear and tear due to no rubbing action. It has less amount surface contact with the surface in rotation.
Thus, tire rotation is the least likely cause of tire wear. So, the option D is correct.
