When the extra work is less than the useful work, the mechanical efficiency of the machine is improved A. More than 0.5 B. Less than 0.5 C. Greater than 1 D. Equal to 0.5

When the extra work is less than the useful work, the mechanical efficiency of the machine is improved A. More than 0.5 B. Less than 0.5 C. Greater than 1 D. Equal to 0.5

A

Judge: is the following statement correct
1. The more useful work is, the higher the mechanical efficiency is
2. The less the extra work, the higher the mechanical efficiency
3. The slower the work, the lower the mechanical efficiency
4. The more total work done, the lower the mechanical efficiency
5. Do the same work, the less extra work, the higher the mechanical efficiency

1. The more useful work is, the higher the mechanical efficiency is
2. The less the extra work, the higher the mechanical efficiency
3. The slower the work, the lower the mechanical efficiency
4. The more total work done, the lower the mechanical efficiency
5. Do the same work, the less extra work, the higher the mechanical efficiency
Efficiency is only related to the ratio of active work to useless work, and has nothing to do with whether the work is long, slow or fast

[physics of senior two] the generator terminal voltage is 220 V and the transmission power is 110 kW. To ensure that the user can obtain at least 100 KW electric power
How many ohms should the resistance of the transmission line not be greater than?
I＝P/U＝110KW/220V＝500A
R = P (consumption) / I * I = 10kW / 500 * 500 = 0.04 ohm
Why is I in the first formula equal to I in the second one? Shouldn't it be N1 / N2 = I2 / I1? The front is the generator terminal voltage, which is not the same as the back voltage?

This is not a transformer
It's just that the 220 voltage is divided by the wire, 20V, to the user, 200V, the generator here, when the power supply is understood

The terminal voltage of alternator is 220 V, the output power is 110 kW, and the resistance of transmission line is 0.3 ohm
1. If the transformer is not used, how much electric power does the user get
2. If the voltage is increased to 2200v by transformer, how much electric power will the user get!

P user = 110k - (110k / 220) ^ 2 * 0.3 = 3.5kw,
After boosting, P user = 110k - (110k / 2200) ^ 2 * 0.3 = 109.25kw;
That's all#