Answer: P = I2R = 0.032 x 1000 =0.9 W
Explanation: The power will be the product of the square of the current and
the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.
Answer:
you know what sounds rlly nice? first have you seent he new corvettes? next u should here a nisan gtr. (like tanner foxes)
Explanation:
Answer:
The axis of motion that is parallel to the spindle axis is always the Z-axis.
Explanation:
Z-Axis Which axis is which depends on the orientation of the spindle.
If the spindle is vertical (Figure 2.1), the Z-axis is vertical. Either the quill or the knee of a vertical spindle mill will move when a Z-axis command is executed.
This is the best answer I could give you, maybe you could show us a pic of the question so we see all the possible choices?
Answer:
a) 75%
b) 82%
Explanation:
Assumptions:
Properties: The density of water 
Conversions:
Analysis:
Note that the bottom of the lake is the reference level. The potential energy of water at the surface becomes gh. Consider that kinetic energy of water at the lake surface & the turbine exit is negligible and the pressure at both locations is the atmospheric pressure and change in the mechanical energy of water between lake surface & turbine exit are:
Then;
gh = 0.491 kJ/kg
= 1559 kW
Therefore; the overall efficiency is:
= 0.75
= 75%
b) mechanical efficiency of the turbine:
thus;
![\eta_{turbine} = \dfrac{\eta_{[turbine- generator]} }{\eta_{generator}} \\ \\ \eta_{turbine} = \dfrac{0.75}{0.92} \\ \\ \eta_{turbine} = 0.82 \\ \\ \eta_{turbine} = 82\%](https://tex.z-dn.net/?f=%5Ceta_%7Bturbine%7D%20%3D%20%5Cdfrac%7B%5Ceta_%7B%5Bturbine-%20generator%5D%7D%20%7D%7B%5Ceta_%7Bgenerator%7D%7D%20%5C%5C%20%5C%5C%20%5Ceta_%7Bturbine%7D%20%3D%20%5Cdfrac%7B0.75%7D%7B0.92%7D%20%5C%5C%20%5C%5C%20%5Ceta_%7Bturbine%7D%20%3D%200.82%20%5C%5C%20%5C%5C%20%5Ceta_%7Bturbine%7D%20%3D%2082%5C%25)
Answer:
The speed is the same at 1.5 m/s while
The work done by the force F is 0.4335 J
Explanation:
Here we have angular acceleration α = v²/r
Force = ma = 2.8 × 1.5²/r₁
and ω₁ = v₁/r₁ = ω₂ = v₁/r₂
The distance moved by the force = 600 - 300 = 300 mm = 0.3 m
If the velocity is constant
The speed is 1.5 m/s while the work done is
2.8 × 1.5²1/(effective radius) ×0.3
r₁ = effective radius
2.8*9.81 = 2.8 × 1.5²/r₁
r₁ = 0.229
The work done by the force = 2.8 × 1.5²*1/r₁ *0.3 = 0.4335 J
