Answer:
20573.67N
Explanation:
Given;
mass (m) of the car = 2130kg
angle of inclination Θ = 15⁰
The normal force (F) on the car is given by
F = mgcosΘ
where g is the acceleration due to gravity.
Taking g as 10
and substituting the values of m and Θ into the equation. We have;
F = 2130 x 10 x cos 15⁰
F = 2130 x 10 x 0.9659
F = 20573.67N
Therefore the normal force on the car is 20573.67N
Answer:
Explanation:
mass of object = m
Mass of planet = M
Radius of planet = R
Height = h
Let the speed of the object as it hits the earth's surface is v.
the value of acceleration due to gravity
g = G M / R^2
where, g is the universal gravitational constant.
Use third equation of motion
where, u is the initial velocity which is equal to zero.
So, 
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .
