The answer to your question is capacitor :)
Explanation & answer:
Given:
Fuel consumption, C = 22 L/h
Specific gravity = 0.8
output power, P = 55 kW
heating value, H = 44,000 kJ/kg
Solution:
Calculate energy intake
E = C*P*H
= (22 L/h) / (3600 s/h) * (1000 mL/L) * (0.8 g/mL) * (44000 kJ/kg)
= (22/3600)*1000*0.8*44000 j/s
= 215111.1 j/s
Calculate output power
P = 55 kW
= 55000 j/s
Efficiency
= output / input
= P/E
=55000 / 215111.1
= 0.2557
= 25.6% to 1 decimal place.
Answer:
fundamental frequency of pipe will be equal to 74 Hz
Explanation:
We have given for a particular organ pipe two adjacent frequency are 296 Hz and 370 Hz
Speed of the sound in air is 343 m/sec
We have to find the fundamental frequency for the pipe
Fundamental frequency will be equal to difference of the two adjacent frequency
So fundamental frequency = 370 - 296 = 74 Hz
So fundamental frequency of pipe will be equal to 74 Hz
One form of Ohm's Law says . . . . . Resistance = Voltage / Current .
R = V / I
R = (12 v) / (0.025 A)
R = (12 / 0.025) (V/I)
<em>R = 480 Ohms</em>
I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)
If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.
If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.