Answer:
The magnitude of the car's acceleration as it slows during braking is 36.81 m/s²
Explanation:
From the question, the given values are as follows:
Initial velocity, u = 90 m/s
final velocity, v = 0 m/s
distance, s = 110 m
acceleration, a = ?
Using the equation of motion, v² = u² + 2as
(90)² + 2 * 110 * a = 0
8100 + 220a = 0
220a = -8100
a = -8100/220
a = -36.81 m/s²
The value for acceleration is negative showing that car is decelerating to a stop. The magnitude of the car's acceleration as it slows during braking is therefore 36.81 m/s²
Answer:
4987N
Explanation:
Step 1:
Data obtained from the question include:
Mass (m) = 0.140 kg
Initial velocity (U) = 28.9 m/s
Time (t) = 1.85 ms = 1.85x10^-3s
Final velocity (V) = 37.0 m/s
Force (F) =?
Step 2:
Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:
F = m(V + U) /t
F = 0.140(37+ 28.9) /1.85x10^-3
F = 9.226/1.85x10^-3
F = 4987N
Therefore, the magnitude of the horizontal force applied is 4987N
Newton showed that IF his theory of universal gravitation is correct, then planets, moons and comets MUST move exactly according to Kepler's laws, and exactly the way that we see them move in the sky.
This demonstration was a spectacular triumph of Newton's theory of gravity, AND of Kepler's laws of planetary motion, AND of the whole heliocentric model of the solar system.
True, think of a rubber band, the more it is stretched the more potential energy that when you let go can be turned into more kinetic energy than if you had less tension on the rubber band.
