Answer:
No, the car will not make it to the top of the hill.
Explanation:
Let ΔX be how long the slope of the hill is, Δx be how far the car will travel along the slope of the hill, Ф be the angle the slope of the hill makes with the horizontal(bottom of the hill), ki be the kinetic energy of the car at the bottom of the hill and vi be the velocity of the car at the bottom of the hill and kf be the kinetic energy of the car when it stop moving at vf.
Since Ф is the angle between the horizontal and the slope, the relationship between the angle and the slope and the height of the hill is given by
sinФ = 12/ΔX
Which gives you the slope as
ΔX = 12/sinФ
Therefore for the car to reach the top of the hill it will have to travel ΔX.
Ignoring friction the total work done is given by
W = ΔK
W = (kf - ki)
Since the car will come to a stop, kf = 0 J
W = -ki
m×g×sinФ×Δx = 1/2×m×vi^2
(9.8)×sinФ×Δx = 1/2×(10)^2
sinФΔx = 5.1
Δx = 5.1/sinФ
ΔX>>Δx Ф ∈ (0° , 90°)
(Note that the maximum angle Ф is 90° because the slope of a hill can never be greater ≥ 90° because that would then mean the car cannot travel uphill.)
Since the car can never travel the distance of the slope, it can never make it to the top of the hill.
Kinetic energy is the energy applied or present in a moving object. According to Newton's second law of motion the magnitude of acceleration of an object is proportional to the magnitude of the net force but inversely proportional to its mass. So the Kinetic Energy of a moving car of small vehicle is greater than the large vehicle if both are applied with the same net force. The greater the Kinetic Energy the longer the stopping distance
Answer:
7kgm/s
Explanation:
Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Let P1A and P1B be the initial momentum of the bodies A and B respectively
Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.
Based on the law:
P1A+P2A = P1B + P2B
Given P1A = 5kgm/s
P2A = 0kgm/s(ball B at rest before collision)
P2A = -2.0kgm/s (negative because it moves in the negative x direction)
P2B = ?
Substituting the values in the equation gives;
5+0 = -2+P2B
5+2 = P2B
P2B = 7kgm/s
High school???
No way
It's work.
I think it is False because as the Gad relajases fuel it doesn’t move as much anymore
