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When you have 12 coils on the output side of a transformer and 3 coils on the input side with 100 volts output what is the input voltage?
25
A machine with a 5-n input force and a 25-n output force has a mechanical advantage of?
Just divide the output force by the input force.Just divide the output force by the input force.Just divide the output force by the input force.Just divide the output force by the input force.
Calculate the power out put of an amplifier that has an input of 15wand a gain of 25dB?
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 Watts x 316.2 = 4.7kW If you want to work in dB, then convert 15 watts to dB: 10 * log |P| = dB = 10*log |15| = 11.76dB so the output is 11.76 + 25 = 36.76dB
What is the machine advantage a 5-n input force and 25-n output force?
The mechanical advantage is calculated by dividing the output force by the input force. In this case, the mechanical advantage would be 25 N (output force) divided by 5 N (input force), resulting in a mechanical advantage of 5.
What is the operation for input numbers 5 10 15 20 25 30 output numbers 4 5 6 7 8 9 10?
The operation appears to involve subtracting 1 from the quotient of each input number divided by 5. Specifically, for each input number ( x ), the output can be calculated as ( \text{output} = \frac{x}{5} + 3 ). For example, for the input 5, the output is ( \frac{5}{5} + 3 = 4 ). This pattern holds for all given input numbers.
What is the efficiency of a ramp if the input work is 96 j and the output work is 24j?
The efficiency of the ramp is 25%. This is calculated by taking the ratio of output work to input work, which in this case is 24 J / 96 J = 0.25, or 25%.
What is the efficiency of a cars engine when heat input is 2000000 joules and the waste heat is 1500000 joules?
25%Take Input - Waste = Differencethen D / I = Effeciencyso, 400000 - 300000 = 100000and 100000 / 400000 = .25 or 25%now ask the teacher next time...Dont answer the question if you are going to get an attitude...
What is the efficiency of an air conditioner if there is a work input of 320 J and a work output of 80 J?
Efficiency = (output/input) x 100 = (80/320) x 100 = 25%
Suppose work input is 25 J and the output distance is 10 m. Factoring in the effect of friction which must be true about output force?
Considering the work input and output distance, to account for the effect of friction, the output force must be greater than the input force due to the work lost to friction. This means that the output force required to move the object the specified distance against friction will be higher than what would be calculated based solely on the given work input and output distance.
Calculate the power out put of an amplifier that has an input of 15 μw and a gain of 25dB?
If you want to work in watts, convert 25dB to a scalling factor: 3dB = 2 x input 10dB = 10 x input 20dB = 100 x input ...25dB = 10 ^ (25/10) = 316.2 x input So the output is 15 micro Watts x 316.2 = (4700)/(10^6) = 4.7 milli watts If you want to work in dB, then convert 15 micro watts to dB: 10 * log |P| = dB = 10*log |15 x 10^6| = -48.2dB ***When you have very small (ie negative) dB, it is often referred to in dBm, or 1/1000 of dB ( 30 dBm = 0 dB) so the output is -18.2dBm + 25 = 6.8dBm, or -23.2dB
What is the efficiency of a street cleaners engine when the heat input is 200 000 joules and the waste heat is 150 000 joules?
The efficiency of the street cleaner's engine is calculated by dividing the useful work output by the total heat input. In this case, the useful work output is the heat input minus the waste heat: 200,000 J - 150,000 J = 50,000 J. Therefore, the efficiency would be 50,000 J (useful work output) / 200,000 J (total heat input) = 0.25 or 25%.
What is the efficiency of a car's engine when he input is 20000 joules and the waste heat is 15000 joules?
The efficiency of the car's engine is calculated by dividing the useful energy output by the energy input. In this case, the useful energy output would be 5000 joules (20000 input - 15000 waste heat), thus the efficiency would be 5000/20000 = 0.25 or 25%.
