Answer: ![\frac{V^{2}}{r}](https://tex.z-dn.net/?f=%5Cfrac%7BV%5E%7B2%7D%7D%7Br%7D)
Explanation:
According to Newton's 2nd Law of motion the force
is proportional to the mass
and acceleration
:
(1)
On the other hand, the equation for the Centripetal force is:
(2)
Where:
is the velocity
is the radius of the circular motion
Making (1) and (2) equal:
(3)
Hence:
This is the expression for the centripetal acceleration
It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).
just swim In water and find your shoe
We can solve the problem by using the first law of thermodynamics:
![\Delta U = Q-W](https://tex.z-dn.net/?f=%20%5CDelta%20U%20%3D%20Q-W%20)
where
is the change in internal energy of the system
is the heat absorbed by the system
is the work done by the system on the surrounding
In this problem, the work done by the system is
![W=-225 kcal=-941.4 kJ](https://tex.z-dn.net/?f=%20W%3D-225%20kcal%3D-941.4%20kJ%20)
with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is
![Q=-5 \cdot 10^2 kJ=-500 kJ](https://tex.z-dn.net/?f=%20Q%3D-5%20%5Ccdot%2010%5E2%20kJ%3D-500%20kJ%20)
with a negative sign as well because it is released by the system.
Therefore, by using the initial equation, we find
![\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ](https://tex.z-dn.net/?f=%20%5CDelta%20U%3DQ-W%3D-500%20kJ%2B941.4%20kJ%3D441.4%20kJ%20)
Answer:
Δu=1300kJ/kg
Explanation:
Energy at the initial state
![p_{1}=200kPa\\t_{1}=300^{o}\\u_{1}=2808.8kJ/kg(tableA-5)](https://tex.z-dn.net/?f=p_%7B1%7D%3D200kPa%5C%5Ct_%7B1%7D%3D300%5E%7Bo%7D%5C%5Cu_%7B1%7D%3D2808.8kJ%2Fkg%28tableA-5%29)
Is saturated vapor at initial pressure we have
![p_{2}=200kPa\\x_{2}=1(stat.vapor)\\v_{2}=0.8858m^3/kg(tableA-5)](https://tex.z-dn.net/?f=p_%7B2%7D%3D200kPa%5C%5Cx_%7B2%7D%3D1%28stat.vapor%29%5C%5Cv_%7B2%7D%3D0.8858m%5E3%2Fkg%28tableA-5%29)
Process 2-3 is a constant volume process
![p_{3}=100kPa\\v_{3}=v_{2}=0.8858m^{3}/kg\\u_{3}=1508.6kJ/kg(tableA-5)](https://tex.z-dn.net/?f=p_%7B3%7D%3D100kPa%5C%5Cv_%7B3%7D%3Dv_%7B2%7D%3D0.8858m%5E%7B3%7D%2Fkg%5C%5Cu_%7B3%7D%3D1508.6kJ%2Fkg%28tableA-5%29)
The overall in internal energy
Δu=u₁-u₃
We replace the values in equation
Δu=u₁-u₃
![=2808.8kJ/kg-1508.6kJ/kg\\=1300kJ/kg](https://tex.z-dn.net/?f=%3D2808.8kJ%2Fkg-1508.6kJ%2Fkg%5C%5C%3D1300kJ%2Fkg)
Δu=1300kJ/kg
Hello friend!!
We know that kinetic energy is the energy possessed due to the motion of the object. And we know if the object is in a fast motion then the temperature would be high, whereas if the object is slow in motion then it will have lower temperature. So we know that the kinetic energy is indirectly related to temperature.From our knowledge we can conclude that HIGHER THE TEMPERATURE, HIGHER THE KINETIC ENERGY and LOWER THE TEMPERATURE, LOWER THE KINETIC ENERGY.
Hence, the answer to your question here is,a.kinetic energy, temperature, and thermal energy increase.
Hope it helps!!All the best!!