Answer:15,883.63 KJ
Explanation:
Given
Power=400 W
Freezing compartment temperature is 273 K
Outside air Temperature=306 K
Time =80 minutes
Energy Required to deliver 400 w power in 80 minutes
E=1920 KJ
and we know COP of refrigerator is given by




Therefore 15,883.63 KJ is removed in 80 minutes
Two thermometers, calibrated in celsius and fahrenheit respectively, are put into a liquid. the reading on the fahrenheit scale is four times the reading on the celsius scale. the temperature of the liquid is:
Resistance = (voltage) / (current)
Resistance = (6.0 v) / (2.0 A)
Resistance = 3.0 ohms
Answer:
346.66 Hz
Explanation:
= Length of string which is unfingered = l
= Length of string which is vibrate when fingered = 
= Unfingered frequency = 260 Hz
= Fingered frequency
Frequency is inversely proportional to length

So,

The frequency of the fingered string is 346.66 Hz