Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
v
Convert the given temperatures from celsius to kelvin since we are dealing with gas.
To convert to kelvin, add 273.15 to the temperature in celsius.
T1 = 22 + 273.15 = 295.15 k
T2 = 4 + 273.15 = 277.15 k
V1 = 0.5 L
Let's find the final volume (V2).
To solve for V2 apply Charles Law formula below:
Explanation:
Given that,
Weight of water = 25 kg
Temperature = 23°C
Weight of mass = 32 kg
Distance = 5 m
(a). We need to calculate the amount of work done on the water
Using formula of work done



The amount of work done on the water is 1568 J.
(b). We need to calculate the internal-energy change of the water
Using formula of internal energy
The change in internal energy of the water equal to the amount of the work done on the water.


The change in internal energy is 1568 J.
(c). We need to calculate the final temperature of the water
Using formula of the change internal energy





The final temperature of the water is 23.01°C.
(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.
The amount of heat is 1568 J.
Hence, This is the required solution.
A force can be considered a push or pull
hope this helps :)