Answer:
![P'_{dynamic} = 72.0\ W](https://tex.z-dn.net/?f=P%27_%7Bdynamic%7D%20%3D%2072.0%5C%20W)
![P'_{leakage} = 15\ W](https://tex.z-dn.net/?f=P%27_%7Bleakage%7D%20%3D%2015%5C%20W)
Solution:
As per the question:
Initial frequency of the processor, f = 2.5 GHz = ![2.5\times 10^{9}\ Hz](https://tex.z-dn.net/?f=2.5%5Ctimes%2010%5E%7B9%7D%5C%20Hz)
Final frequency attained by the processor, f' = ![3.0\times 10^{9}\ Hz](https://tex.z-dn.net/?f=3.0%5Ctimes%2010%5E%7B9%7D%5C%20Hz)
Initial dynamic power, ![P_{dynamic} = 60\ W](https://tex.z-dn.net/?f=P_%7Bdynamic%7D%20%3D%2060%5C%20W)
Leakage Power, ![P_{leakage} = 15\ W](https://tex.z-dn.net/?f=P_%7Bleakage%7D%20%3D%2015%5C%20W)
Now,
To calculate the Dynamic power,
and leakage power consumed by the processor:
We know that the dynamic power is in direct proportion to the frequency of the processor:
∝ frequency, f
Thus we can write:
![\frac{P'_{dynamic}}{P_{dynamic}} = \frac{f'}{f}](https://tex.z-dn.net/?f=%5Cfrac%7BP%27_%7Bdynamic%7D%7D%7BP_%7Bdynamic%7D%7D%20%3D%20%5Cfrac%7Bf%27%7D%7Bf%7D)
![P'_{dynamic} = \frac{f'}{f}\times P_{dynamic](https://tex.z-dn.net/?f=P%27_%7Bdynamic%7D%20%3D%20%5Cfrac%7Bf%27%7D%7Bf%7D%5Ctimes%20P_%7Bdynamic)
Substituting the appropriate values in the above expression:
![P'_{dynamic} = \frac{3.0\times 10^{9}}{2.5\times 10^{9}}\times 60 = 72.0\ W](https://tex.z-dn.net/?f=P%27_%7Bdynamic%7D%20%3D%20%5Cfrac%7B3.0%5Ctimes%2010%5E%7B9%7D%7D%7B2.5%5Ctimes%2010%5E%7B9%7D%7D%5Ctimes%2060%20%3D%2072.0%5C%20W)
Now,
We know that the leakage power does not depend on the frequency of the processor and hence remains same.
Thus
![P'_{leakage} = P_{leakage} =15\ W](https://tex.z-dn.net/?f=P%27_%7Bleakage%7D%20%3D%20P_%7Bleakage%7D%20%3D15%5C%20W)
I think it is equal for your question
B is the answer because of its throwing the other 2 have potential energy
Answer:
they become ionized and change to ions
Answer: Potential energy to kinetic energy
Explanation:
Potential energy is the energy by virtue of its position
P.E = mgh
While
Kinetic energy is the energy by virtue of motion.
K.E = 0.5mv^2
At the top of the hill the boulder has maximum height and thereby possess maximum P.E, as it rolls down the hill the height reduces and velocity increases, so the P.E reduces and K.E increases(P.E is directly proportional to height while K.E is directly proportional to velocity). So as the boulder rolls down the hill its potential energy is converted to kinetic energy (according to the rule of conservation of energy).