The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
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Newton´s thrid law of motion states that:
Every action has an equal and opposite reaction.
c. equal, opposite.
I believe it’s A, i could be wrong tho 3
Answer:
and 
Explanation:
The wavelength of a visible light is 727.3 nm.
The formula is as follows :
f is the frequency of the visible light
Energy of a photon is given by :
E = hf, h is Planck's constant
Red color has a frequency of 
and energy per photon is 
.
Answer:
1. Kinetic Energy = 0.0161 Joules
2. Height = 0.0137m
Explanation:
Given
Length of Rod, l = 0.64m
Mass, m = 120g = 0.12kg
Angular speed, w = 1.40 rad/s
a.
Calculating the Rod's kinetic energy
This is calculated by
Kinetic Energy = ½Iw²
Where I = rotational inertia of the rod about an axis.
This is calculated as follows;
I = Icm + mh²
I = ImL² + m(L/2)²
I = 1/12 * 0.12 * 0.64² + 0.12 * (0.64/2)²
I = 0.016384 kgm²
By substituton
KE = ½Iw² becomes
KE = ½ * 0.016384 * 1.40²
KE = 0.01605632J
KE = 0.0161 Joules
2. Using the total conservation of momentum;
K + U = Kf + V
Where K = Initial Kinetic Energy of the rod at lowest point.
U = Initial gravitational potential energy of the rod at lowest point
Kf = Final Kinetic Energy of the rod at maximum height = 0 J
V = Final gravitational potential energy of the rod at maximum height
So, K + U = Kf + V become
K + U = 0 + V
K + U = V
K = V - U = mgh
substitute 0.01605632J for K
0.01605632J = mgh
h = 0.01605632J/mg
h = 0.01605632J/(0.12 * 9.8)
h = 0.013653333333333
h = 0.0137m
