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:
Building codes are generally intended to be applied by architects, engineers, interior designers, constructors and regulators but are also used for various purposes by safety inspectors, environmental scientists, real estate developers, subcontractors, manufacturers of building products and materials, insurance companies, facility managers, tenants, and others.
Explanation:
<em>Engineering</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>profession</em><em> </em><em>which</em><em> </em><em>deals </em><em>with</em><em> </em><em>building</em><em> </em><em>the</em><em> </em><em>structures</em><em> </em><em>like</em><em> </em><em>bridge</em><em>,</em><em>house</em><em>,</em><em>roads</em><em> </em><em>etc</em><em>.</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>
The question is not complete. We are supposed to find the average value of v_o.
Answer:
v_o,avg = 0.441V
Explanation:
Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;
3/(T/4) = 0.7/t1
So, 12/T = 0.7/t1
So, t1 = 0.7T/12
t1 = 0.0583 T
Also, from symmetry of triangles,
t2 = T/2 - t1
So, t2 = T/2 - 0.0583 T
t2 = 0.4417T
Average of voltage output is;
v_o,avg = (1/T) x Area under small triangle
v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)
v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)
v_o,avg = (1/T) x 2.3 x 0.1917T
T will cancel out to give;
v_o,avg = 0.441V
Answer:
Technician A is correct
Explanation:
The neutral safety switch when bad must be replaced and not adjusted as suggested by technician B because if the neutral safety switch is bad the Engine might not crank when put in neutral but it will crank when put in park and this is very bad for the life of the Engine it is better to replace it. A test for a bad/faulty neutral safety switch will be required to ascertain the level of damage it might cause to the Engine and prompt replacement is essential as well. because the neutral helps to prevent the car from starting when it is already engaged in a gear position therefore protecting the car from sudden collisions
