Answer:
828 kg/m³ or 0.828 g/cm³
Explanation:
Applying,
D = m/V............. Equation 1
Where D = density of the liquid, m = mass of the liquid, V = volume of the liquid.
From the question,
Given: m = 77 g , V = 93 cm³
Substitute these values into equation 1
D = 77/93
D = 0.828 g/cm³
Converting to kg/m³
D = 828 kg/m³
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:
<h2>b) 4230 J
</h2>
Explanation:
Step one:
given data
extension= 40cm
Spring constant K= 52.9N/cm
Step two:
Required
the Kinetic Energy KE
the expression to find the kinetic energy is
KE= 1/2ke^2
substituting our data we have
KE= 1/2*52.9*40^2
KE=0.5*52.9*1600
KE= 42320Joules
<u>The answer is b) 4230 J
</u>