Answer:
The ice melts mass is:
Explanation:
Kinetic Energy
Heat gained by ice= mass(g) x 80 cal
( 1 cal = 4.184 *10^7er or g cm^2/ sec^2)
Assuming no loss in heat, in the motion so both continue with temperature 0~C
To find so the mass (gm) of ice melted
Answer:
W=1705.2 J
Explanation:
Given that
mass ,m= 60 kg
Acceleration due to gravity ,g= 9.8 m/s²
Height ,h= 2.9 m
As we know that work done by a force given as
W = F . d
F=force
d=Displacement
W=work done by force
Now by putting the values
F= m g (Acting downward )
d= h (Upward)
W= m g h ( work done against the force)
W= 60 x 9.8 x 2.9 J
W=1705.2 J
Therefore the answer will be 1705.2 J.
Rhythms that occur faster and slower than the beat are b.<span>not synchronized with the time signature. The synchronization follows the same beat or rhythm. If the time signature say is lower than the original, then the rhythm should be faster. Otherwise, the rhythm is slower than the original one.</span>
It means that if you had a cubic meter of water it would weigh 1000 kilograms
Answer:
- The work made by the gas is 7475.69 joules
- The heat absorbed is 7475.69 joules
Explanation:
<h3>
Work</h3>
We know that the differential work made by the gas its defined as:
We can solve this by integration:
but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law
This give us
As n, R and T are constants
![\Delta W= \ n \ R \ T \left [ ln (V) \right ]^{v_2}_{v_1}](https://tex.z-dn.net/?f=%20%5CDelta%20W%3D%20%5C%20n%20%5C%20R%20%5C%20T%20%20%5Cleft%20%5B%20ln%20%28V%29%20%5Cright%20%5D%5E%7Bv_2%7D_%7Bv_1%7D%20)
But the volume is:
Now, lets use the value from the problem.
The temperature its:
The ideal gas constant:
So:
<h3>Heat</h3>
We know that, for an ideal gas, the energy is:
where 
its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.
By the first law of thermodynamics, we know
where 
is the Work made by the gas (please, be careful with this sign convention, its not always the same.)
So:
