Answer:
(d) all of the above
Explanation:
before cutting the screw threads the operator should lubricate all of the machine parts given in the option that is lead screw and gearbox , the ways and the cross slide and the carriage and half-nuts. we should use lubrication because it reduces the overall system friction and if friction is reduced then heat generated due to friction is also decreases which is beneficial
so option (D) will be correct because we need lubricate in all the given parts
On highways, the far left lane is usually the<u> fastest</u> moving traffic.
Answer: Option D.
<u>Explanation:</u>
For the most part, the right lane of a freeway is for entering and leaving the traffic stream. It is an arranging path, for use toward the start and end of your interstate run. The center paths are for through traffic, and the left path is for passing. On the off chance that you are not passing somebody, try not to be driving in the left path.
Regular practice and most law on United States expressways is that the left path is saved for passing and quicker moving traffic, and that traffic utilizing the left path must respect traffic wishing to surpass.
Answer:
(4.5125 * 10^-3 kg.m^2)ω_A^2
Explanation:
solution:
Moments of inertia:
I = mk^2
Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2
Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2
Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2
Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.
r_A=50 mm
r_1 =100 mm
r_2 =50 mm
r_C=150 mm
At the contact point between gears A and B,
r_1*ω_b = r_A*ω_A
ω_b = r_A/r_1*ω_A
= 0.5ω_A
At the contact point between gear B and C.
At the contact point between gears A and B,
r_C*ω_C = r_2*ω_B
ω_C = r_2/r_C*ω_B
= 0.1667ω_A
kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2
=(4.5125 * 10^-3 kg.m^2)ω_A^2
Answer:
COP_max = 18.69
Explanation:
We are given;
Heated space temperature; T_H = 26°C = 273K + 26 = 299K
Temperature at which heat is extracted; T_L = 10°C = 273 + 10 = 283K
Now the Coefficient of performance (COP) of a heat pump will be a maximum when the heat pump operates in a reversible manner. The COP of a reversible heat pump depends on the temperature limits in the cycle only and is determined by the formula;
COP_max = 1/(1 - (T_L/T_H))
Thus,
COP_max = 1/(1 - (283/299))
COP_max = 1/(1 - 0.9465)
COP_max = 1/0.0535 = 18.69
Answer:
when 5% excess air is supplied, moles of air supplied/moles of fuel = 
Explanation:
Equivalence ratio = 0.6
Equivalence ratio = Actual air to fuel ratio (AAFR)/ stoichiometric air to fuel ratio SAFR
combustion reaction of propane is
From above reaction, 1 mole of propane, from the reaction, 5 moles of oxygen required,
we know that air contains 21% O_2 and 79% N_2,
Therefore, moles of air based on stoichiometry 
Theoretical air to fuel ratio 
Given
Actual Air Fuel Ratio 
when 5% excess air is supplied, moles of air supplied/moles of fuel =
