Answer:
is time required to heat to boiling point form initial temperature.
Explanation:
Given:
initial temperature of water, 
time taken to vapourize half a liter of water, 
desity of water, 
So, the givne mass of water, 
enthalpy of vaporization of water, 
specific heat of water, 
Amount of heat required to raise the temperature of given water mass to 100°C:
Now the amount of heat required to vaporize 0.5 kg of water:
where:
mass of water vaporized due to boiling
Now the power rating of the boiler:
Now the time required to heat to boiling point form initial temperature:
Answer:
8.13secs
Explanation:
From the question weal are given
Height H =324m
Required
time it takes to drop t
Using the equation of motion
H = ut + 1/2gt²
Substitute the given values
324 = 0(t)+1/2(9.8)t²
324 = 1/2(9.8)t²
324 = 4.9t²
t² =324/4.9
t² = 66.12
t = √66.12
t = 8.13secs
Hence the time taken to drop is 8.13secs
magnitude of the net force = mass x acceleraton
= 22 x 2.3
=50.6 N
Answer:
K = -½U
Explanation:
From Newton's law of gravitation, the formula for gravitational potential energy is;
U = -GMm/R
Where,
G is gravitational constant
M and m are the two masses exerting the forces
R is the distance between the two objects
Now, in the question, we are given that kinetic energy is;
K = GMm/2R
Re-rranging, we have;
K = ½(GMm/R)
Comparing the equation of kinetic energy to that of potential energy, we can derive that gravitational kinetic energy can be expressed in terms of potential energy as;
K = -½U
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