Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
<em>Q = mcΔT</em>
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
Answer : The final temperature of gas is 266.12 K
Explanation :
According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.
The formula will be:

or,

As per question the formula will be:
.........(1)
where,
= Joule-Thomson coefficient of the gas = 
= initial temperature = 
= final temperature = ?
= initial pressure = 200.0 atm
= final pressure = 0.95 atm
Now put all the given values in the above equation 1, we get:


Therefore, the final temperature of gas is 266.12 K
Neither set of choices is correct.
If the distance is tripled, then the forces decrease to
1/9 Fg. and. 1/9 Fe.
Note. When the objects are charged, the gravitational force Fg can almost always be ignored, since Fe is like 10^40 greater when the quantities of mass and charge are similar.
Answer:
It is larger than the Sun.
Explanation:
Brainliest pls
:3