2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s

a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)


2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
The correct answer is amplitude
Answer : The volume of a sample of 4.00 mol of copper is 
Explanation :
First we have to calculate the mass of copper.


Now we have to calculate the volume of copper.
Formula used :

Now put all the given values in this formula, we get:


Conversion used :

Therefore, the volume of a sample of 4.00 mol of copper is 
Density = (mass) divided by (volume)
We know the mass (2.5 g). We need to find the volume.
The penny is a very short cylinder.
The volume of a cylinder is (π · radius² · height).
The penny's radius is 1/2 of its diameter = 9.775 mm.
The 'height' of the cylinder is the penny's thickness = 1.55 mm.
Volume = (π) (9.775 mm)² (1.55 mm)
= (π) (95.55 mm²) (1.55 mm)
= (π) (148.1 mm³)
= 465.3 mm³
We know the volume now. So we could state the density of the penny,
but nobody will understand what we have. Here it is:
mass/volume = 2.5 g / 465.3 mm³ = 0.0054 g/mm³ .
Nobody every talks about density in units of ' gram/(millimeter)³ ' .
It's always ' gram / (centimeter)³ '.
So we have to convert our number for the volume.
(0.0054 g/mm³) x (10 mm / cm)³
= (0.0054 x 1,000) g/cm³
= 5.37 g/cm³ .
This isn't actually very close to what the US mint says for the density
of a penny, but it's in a much better ball park than 0.0054 was.