Answer:
A. 79.3 keV
Explanation:
Because the procedure involves many steps for its resolution and it works faster on paper and pencil, the detailed solution of this exercise is attached as a scanned image of the procedure for review.
In the procedure, the initial values of the problem and the replacement of these values with the correct formulas for this process are taken into account.
THe loss of 2 protons and 2 neutrons (also called a helium nucleus) is defined as alpha decay.
Answer:
1.04%
Explanation:
Given that,
The power of drill = 3.5 kW = 3500 W
Transferred kinetic energy = 5000 kJ during 15 seconds of use.
We need to find the percentage efficiency of the drill. It can be given by :

Where
Po and Pi are output and input powers.

So,

So, the percentage efficiency of the drill is 1.04%.
depending on the object the force pressing in on the object are equal with the quantity of the object.