Answer: See below
Explanation:
<u>Given:</u>
Mass of the Mercedes-Benz (m) = 1700 kg
Inclination of the road (θ) = 15.0
<em>The free body diagram is shown in figure attached below</em>
<em />
a) The normal force is equal to the cos component of the weight of the car.

b) The static force will be equal to the weight's sin component.

Answer:
49 N
Explanation:
In order to move the box at constant speed, the acceleration of the box must be zero (a=0): this means, according to Newton's second law,
F = ma
that the net force acting on the box, F, must be zero as well.
Here there are two forces acting on the box in the horizontal direction while it is moving:
- The force of push applied by the guy, F
- The frictional force, 
For an object moving on a flat surface, the frictional force is given by

where
is the coefficient of friction
m is the mass of the box
g is the acceleration of gravity
So the equation of the forces becomes

And substituting:

We find the force that must be applied by the guy:

Answer:
The gravitational potential energy of a system is -3/2 (GmE)(m)/RE
Explanation:
Given
mE = Mass of Earth
RE = Radius of Earth
G = Gravitational Constant
Let p = The mass density of the earth is
p = M/(4/3πRE³)
p = 3M/4πRE³
Taking for instance,a very thin spherical shell in the earth;
Let r = radius
dr = thickness
Its volume is given by;
dV = 4πr²dr
Since mass = density* volume;
It's mass would be
dm = p * 4πr²dr
The gravitational potential at the center due would equal;
dV = -Gdm/r
Substitute (p * 4πr²dr) for dm
dV = -G(p * 4πr²dr)/r
dV = -G(p * 4πrdr)
The gravitational potential at the center of the earth would equal;
V = ∫dV
V = ∫ -G(p * 4πrdr) {RE,0}
V = -4πGp∫rdr {RE,0}
V = -4πGp (r²/2) {RE,0}
V = -4πGp{RE²/2)
V = -4Gπ * 3M/4πRE³ * RE²/2
V = -3/2 GmE/RE
The gravitational potential energy of the system of the earth and the brick at the center equals
U = Vm
U = -3/2 GmE/RE * m
U = -3/2 (GmE)(m)/RE
Answer:
I think c buti don't think I'm correct