Answer:
a) 4.7 kΩ, +/- 5%
b) 2.0 MΩ, +/- 20%
Explanation:
a) If the resistor has the following combination of color bands:
1) Yellow = 1st digit = 4
2) Violet = 2nd digit = 7
3) Red = multiplier = 10e2
4) Gold = tolerance = +/- 5%
this means that the resistor has 4700 Ω (or 4.7 kΩ), with 5% tolerance.
b) Repeating the process for the following combination of color bands:
1) Red = 1st digit = 2
2) Black = 2nd digit = 0
3) Green = multiplier = 10e5
4) Nothing = tolerance = +/- 20%
This combination represents to a resistor of 2*10⁶ Ω (or 2.0 MΩ), with +/- 20% tolerance.
Technician A is correct. Technician B is wrong because a gear's transmission is used to increase or decrease torque.
The relation torque is relying on multiplying the circumferential detail with the resource of the usage of the radius; massive gears experience a greater amount of torque, at the same time as smaller gears experience a great deal much less torque. Similarly, the torque ratio is equal to the ratio of the gears' radii. A gear's transmission torque modifications as it will boom or decreases speed. Commonly, with the resource of the usage of lowering the speed, a small torque on the doorway issue is transferred as a massive torque at the output issue. The calculation of torque is quantified with the resource of the usage of an extensive form of teeth.
Learn more about the torque at brainly.com/question/28220969
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Answer:
Q=67.95 W
T=119.83°C
Explanation:
Given that
For air
Cp = 1.005 kJ/kg·°C
T= 20°C
V=0.6 m³/s
P= 95 KPa
We know that for air
P V = m' R T
95 x 0.6 = m x 0.287 x 293
m=0.677 kg/s
For gas
Cp = 1.10 kJ/kg·°C
m'=0.95 kg/s
Ti=160°C ,To= 95°C
Heat loose by gas = Heat gain by air
[m Cp ΔT] for air =[m Cp ΔT] for gas
by putting the values
0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )
T=119.83°C
T is the exit temperature of the air.
Heat transfer
Q=[m Cp ΔT] for gas
Q=0.95 x 1.1 x ( 160 -95 )
Q=67.95 W
The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.
