You know that when the displacement is equal to the amplitude (A), the velocity is zero, which implies that the kinetic energy (KE) is zeero, so the total mechanical energy (ME) is the potential energy (PE).
And you know that the potential energy, PE, is [ 1/2 ] k (x^2)
Then, use x = A, to calculate the PE in the point where ME = PE.
ME = PE = [1/2] k (A)^2.
At half of the amplitude, x = A/2 => PE = [ 1/2] k (A/2)^2
=> PE = [1/4] { [1/2]k(A)^2 } = .[1/4] ME
So, if PE is 1/4 of ME, KE is 3/4 of ME.
And the answer is 3/4
the friction force provided by the brakes is 30000 N.
<h3>What is friction force?</h3>
Friction force is the force that opposes the motion between two bodies in contact.
To calculate the average friction force provided by the brakes, we apply the formula below.
Formula:
- K.E = F'd............. Equation 1
Where:
- K.E = Kinetic energy of the train
- F' = Friction force provided by the brakes
- d = distance
Make F' the subject of the equation
- F' = K.E/d............ Equation 2
From the question,
Given:
Substitute these values into equation 2
- F' = (8.1 ×10⁶)/270
- F' = 30000 N
Hence, the friction force provided by the brakes is 30000 N
Learn more about friction force here: brainly.com/question/13680415
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sorry if this is wrong, i hope i helped
Answer:
Magnitude of the force on proton = F = 1.1085 × 10^-15 N
Explanation:
Charge on proton = q = 1.60 × 10^-19 C
Velocity of proton = V = 4.0 × 10^4 m/s
Magnetic field = B = 0.20 T
Angle between V and B = θ = 60
We know that,
F = qVBsin θ = (1.60 × 10^-19)( 4.0 × 10^4)( 0.20)sin(60)
F = 1.1085 × 10^-15 N
Answer:3.31m/s², to the right
Explanation:
According to the law of conservation of momentum of a body, change in momentum of bodies before collision is equal to the change in momentum after collision.
Momentum = mass × velocity
M1 and M2 be the masses of the first and second skaters respectively
Let u1 and u2 be the velocities of the first and second skaters respectively.
v be their common velocity after collision
M1 = 77kg M2 = 66kg u1 = 4m/s² u2 = 2.5m/s²
According to the law we have
M1u1 + M2u2 = (M1+M2)v
77(4) + 66(2.5) = (77+66)v
308 + 165 = 143v
V = 473/143
V = 3.31m/s²
Their velocity after collision will become 3.31m/s²
They will both move towards the right after collision because the mass of the body moving to the right is higher than the other mass and the mass is also moving at a higher velocity than the other.