Answer:
Explanation:
Given
acceleration is given by
where 
Also acceleration is given by
at 
when air drag is neglected maximum height reached is
Answer:
The solution(s) are in order with respect to the attachments
Joules ; 5. Adding the same amount of heat to two different objects will produce the same increase in temperature ; 2. Same speed in both ; 2. A
Explanation:
Diagram 1 ( Liquid Nitrogen ) : So as you can see, we want our units in Joules here, and can therefore multiply the mass of gaseous nitrogen and the latent heat of liquid nitrogen, to cancel the units kg, and receive our solution - in terms of Joules. Let's do it.
q ( energy removed ) = mass of nitrogen 
latent heat of liquid nitrogen,
q = 1.3 kg 2.01 10⁵ J / kg = 
= 
= 
= 
Joules = 
kiloJoules = 2.613 10⁵Joules is the energy that must be removed
Diagram 2 : The same amount of heat does not necessarily mean the same increase in temperature for two different objects. The increase in temperature depends on the specific heat capacity of the substance. Therefore your solution is 5 ) Adding the same amount of heat to two different objects will produce the same increase in temperature.
Diagram 3 : The temperatures in both glasses are the same, and hence the molecules have the same average speed. Therefore your solution is 2 ) Same speed in both.
Diagram 4 : Glass A has more water molecules, and hence has more thermal energy. Your solution is 2 ) A.
In stars more massive than the sun, the core temperature is hotter, which allows for fusion of more complex elements.
Most of the fusion occurs in the core.
In stars more massive than the sun, fusion continues through Deuterium, Carbon, and finally reaching iron/nickel.
Up to this point, the fusion reaction was endothermic, which means that the energy expended to produce the fusion reaction was exceeded by the energy produced in the reaction.
Fusion past iron is exothermic, and therefore the star will be able to survive by fusing elements heavier than iron.
After the core is almost entirely iron, the star is no longer in the Main Sequence.
So, fusion in stars more massive than the sun continue fusing until the core is almost entirely <em>iron</em>.
Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
