r = radius of the circle traveled by the particle = 76 cm = 0.76 m
T = time period of revolution for the particle = 4.5 s
w = angular velocity of the particle
angular velocity of the particle is given as
w = 2π/T
inserting the values
w = 2 (3.14)/4.5
w = 1.4 rad/s
a = centripetal acceleration of the particle in the circle
centripetal acceleration is given as
a = r w²
inserting the values
a = (0.76) (1.4)²
a = 1.5 m/s²
The answer is position 3, because it is at its lowest point.
Potential Energy is “stored energy.” It is energy that is ready to be converted or released as another type of energy. We most often think of potential energy as gravitational potential energy. When objects are higher up, they are ready to fall back down. When you stretch an object and it has a tendency to return to its original shape, it is said to have elastic potential energy. Chemical potential energy is the stored energy in a substance’s chemical structure that can be released in a chemical reaction or as heat.
Explanation:
Total mass=100+10=110
Total weight=mass×gravitational field strength
=110×10
=1100N
Work done=force×distance
=1100×10
=11000J
<em>Please mark me as brainliest if this helped you!</em>
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)
Learn more about kinetic friction here-
brainly.com/question/13754413
#SPJ4
By the use of a Accumulators