the mean of the data set is 6ml
Kinetic energy = (1/2) (mass) x (speed)²
At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules
At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules
The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.
That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
Answer:
31ohms
Explanation:
in a series u add all the ohms together
Answer:
15009
Explanation:
PE = mgh
PE = 61.2(9.81)(10 * 2.50)
PE = 15009.3
To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.
By definition we know that the change in entropy is given by

Where,
Q = Heat transfer
T = Temperature
On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

According to the data given we have to,




PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is



On the other hand,



The total change of entropy would be,



Since
the heat engine is not reversible.
PART B)
Work done by heat engine is given by



Therefore the work in the system is 100000Btu