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
Answer: The gravitational potential energy is
8.74 ×10^8Joules
Explanation:
To find the gravitational potential energy between ANY two objects that have mass, such as the Earth and the moon, you need to know the universal gravitational constant (G), the mass of the two objects (M and m), and the center to center distance between them (r).
The equation for determining gravitational potential energy Ug = - GMm/r
Where G = gravitational constant 6.67×10^-11
r = radius of the earth = 6.378×10^6m
M = mass of the earth = 5.97 ×10^24kg
m = mass of object = 14kg
Substituting into the equation
Ug = [( 6.67×10^11)(5.97×10^24)(14)] / (6.378×10^6)
Ug = (5.575×10^15)/(6.378×10^6)
Ug = 8.74 ×10^8 Joules
Answer:
I don't exactly understand this but the child should hear a beep lower than 500 hz when he/she moves the car further away. If you don't find my answer appealing sorry but I don't do physics.
Explanation:
Answer:
W = 100000 J = 100 KJ
Explanation:
Here we will use the most basic and general formula of work, which is as follows:
where,
W = Work Done = ?
F = Force Required = 200 N
d = Length of Track = 500 m
Therefore,
<u>W = 100000 J = 100 KJ</u>
Answer:
Explanation:
We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.
1 m =100 cm
Surface area =S=
We have to find the charge Q on the positive plates of the capacitor.
V=Initial voltage between plates
d=Initial distance between plates
Initial Capacitance of capacitor
Capacitance of capacitor after moving plates
Potential difference between plates after moving
Hence, the charge on positive plate of capacitor=