The kinetic energy of the small ball before the collision is
KE = (1/2) (mass) (speed)²
= (1/2) (2 kg) (1.5 m/s)
= (1 kg) (2.25 m²/s²)
= 2.25 joules.
Now is a good time to review the Law of Conservation of Energy:
Energy is never created or destroyed.
If it seems that some energy disappeared,
it actually had to go somewhere.
And if it seems like some energy magically appeared,
it actually had to come from somewhere.
The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.
Answer:
C. 5.6 × 10^11 N/C
Explanation:
The electric field 
at a distance 
from a charge 
is given by
where 
is the coulomb's constant.
Now, in our case
;
therefore,
which is choice C from the options given<em> (at least it resembles it).</em>
Answer:
The final temperature of the gas is <em>114.53°C</em>.
Explanation:
Firstly, we calculate the change in internal energy, ΔU from the first law of thermodynamics:
ΔU=Q - W
ΔU = 1180 J - 2020 J = -840 J
Secondly, from the ideal gas law, we calculate the final temperature of the gas, using the change in internal energy:
Then we make the final temperature, T₂, subject of the formula:
Therefore the final temperature of the gas, T₂, is 114.53°C.
But even more pain on pain and then pain and pain ya feel me and even more pain okay and yes more pain
Answer:
(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%
Explanation:
Using the formula that relate heat and work from the thermodynamic theory as:
solving to Q_out we get:
this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: 
where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:
