Answer:
he kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.
Explanation:
In these semicircular sections the skaters slide from one side to the other, in the downward path their kinetic energy increases and their potential energy decreases; When it leaves the ramp and is in the air, the kinetic energy decreases rapidly, up to the point of maximum height where the kinetic energy is zero.
Consequently, the kinetic energy increases on the descent, being maximum at the lowest point of the trajectory.
A chart that organizes all elements by their atomic number.
Answer:
Explanation:
Given that,
Spring constant k=200N/m
Compression x = 15cm = 0.15m
Attached mass m =2kg
Coefficient of kinetic friction uk= 0.2
The energy in the spring is given as
U =½kx²
U = ½ × 200 × 0.15²
U = 2.25J
Force in the spring is given by Hooke's law
F = ke
F = 200×0.15
F = 30N
The weight of body which is equal to the normal is give as
W = mg
W = 2 × 9.81
W = 19.62N
W = N = 19.62 Newton's 2nd Law
From law of friction,
Fr = uk•N
Fr = 0.2 × 19.62
Fr = 3.924
Using newton second law again
Fnet = F - Fr
Fnet = 30 - 3.924
Fnet = 26.076
Work done by net force is given as
W = Fnet × d
W = 26.076d
Then, the work done by this net force is equal to the energy in the spring
W = U
26.076d = 2.25
d = 2.25/26.076
d = 0.0863m
Which is 8.63cm
So the box will slide 8.63cm before stopping
I believe it’s false, I think I had this before
Answer:
The impulse experienced by the car is 52,500 kg.m/s.
Explanation:
Given;
mass of the car, m = 1500 kg
initial velocity of the car, u = 55 m/s
final velocity of the car, v = 90 m/s
The impulse experienced by the car is the change in linear momentum, calculated as follows;
J = ΔP = mv - mu
ΔP = m(v - u)
ΔP = 1500(90 - 55)
ΔP = 52,500 kg.m/s
Therefore, the impulse experienced by the car is 52,500 kg.m/s.
