Answer:
0.4 g/cm^3
Explanation:
The density of an object can be found using the following formula.
d= m/v
where m is the mass and v is the volume.
The mass of the metal is 6 grams and the volume is 15 centimeters^3
m=6 g
v= 15 cm^3
Substitute these into the formula.
d= 6 g/ 15 cm^3
Divide 6 g by 15 cm^3 (6/15=0.4)
d= 0.4 g/ cm^3
The density of the metal is 0.4 grams per cubic centimeter.
-- There's a force of 240N pushing her backwards.
-- She's maintaining a steady speed (of 2.5 m/s) .
-- In order to maintain a steady speed (no acceleration),
the forces on her must be balanced. So she's maintaining
a steady force of 240N forward.
-- Every time she moves 1 m forward, she does work of
(force) x (distance) = 240 joules.
-- She moves 2.5 meters forward every second.
So she's doing (240 x 2.5) = 600 joules of work every second.
-- 600 joules per second = 600 watts .
a) the number of protons is
more than the electrons
b) 
Explanation:
The net electric charge on the ball is

This electric charge is given by the algebraic sum of the charge of the protons and of the charge of the electrons.
The charge of one proton is:

While the charge of one electron is

So the net charge on the metal ball will be given by

where
is the number of protons
is the number of electrons
So we find:

This means that the number of protons is
more than the electrons.
b)
In this case, we want to make the ball neautral, so we have to remove a net charge of Q' such that the new charge is zero:

This implies that the charge that we must remove is

To do that (and to make the ball losing mass at the same time), we have to remove protons, since they have positive charge.
The number of protons that must be removed is:

The mass of one proton is

Therefore, the total mass that must be removed from the ball is

The first law of thermodynamics says that the variation of internal energy of a system is given by:

where Q is the heat delivered by the system, while W is the work done on the system.
We must be careful with the signs here. The sign convention generally used is:
Q positive = Q absorbed by the system
Q negative = Q delivered by the system
W positive = W done on the system
W negative = W done by the system
So, in our problem, the heat is negative because it is releaed by the system:
Q=-1275 J
while the work is positive because it is performed by the surrounding on the system:
W=+855 J
So, the variation of internal energy of the system is