Answer:
A change of one degree Celsius = a change of one Kelvin, but a Celsius temperature is never equal to a Kelvin temperature. A change of 1 degree Fahrenheit equals a change of 5/9 = 0.56 degrees Celsius. To convert a Fahrenheit temperature to Celsius, subtract 32 and multiply by 5/9.
Explanation:
<span>Wave energy is an idea that Robert Hutchings Goddard introduced in his “Further Developments” to his research "A Method of Reaching Extreme Altitudes" </span>
Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - 
where G is the gravitational force constant.
From the above we see that F ∝ mm' and 
Let the orbital radius of planet A is 
= r and mass of planet is 
.
Let the mass of central star is m .
Hence the gravitational force for planet A is 
For planet B the orbital radius 
and mass 
Hence the gravitational force 
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)
Hence the ratio is 
[ ans]
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
