Is there a picture of the car you can attach? That would help me to be able to answer the question :-)
Answer:
In statics condition and if is a body is over plane surface N = mg
In all cases which a body is on a ramp force N will be smaller than mg, and depending on the level of the ramp (bigger the ramp smaller force N)
Explanation:
See annex (Figure 1)
We can see with the help of force diagram the relation between N force and mg
Lets look first the body on the ramp assuming there is not movement
The sum of forces alog y axis = 0
mg* cos∠ BAC = N (normal force)
So N is always a fuction of mg and of the cosine of the inclination angle, and the cosine of angle varies from value 1 when the angle is 0 (look the case where the block is on the surface without inclination N = mg ).
And as the inclination increase the value of cosine will become smaller and so will N force which is directly proportional of cosine of the inclination angle
Therefore we can say that force N is always minor than mg unless in extreme case where the block is over the surface in which case N takes its maximum value N = mg
Answer:
The height of the hill is, h = 38.42 m
Explanation:
Given,
The horizontal velocity of the soccer ball, Vx = 15 m/s
The range of the soccer ball, s = 42 m
The projectile projected from a height is given by the formula
S = Vx [Vy + √(Vy² + 2gh)] / g
Therefore,
h = S²g/2Vx² (Since Vy = 0)
Substituting the values
h = 42² x 9.8/ (2 x 15²)
= 38.42 m
Hence, the height of the hill is, h = 38.42 m
Answer:

Explanation:
Given data
Speed of jet Vjet=1190 km/h
Speed of prop driven Vprop=595 km/h
Height of jet 7.5 km
Height of prop driven transport 3.8 km
Density of Air at height 10 km p7.8=0.53 kg/m³
Density of air at height 3.8 km p3.8=0.74 kg/m³
The drag force is given by:

The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:
