I’m pretty sure it’s circuit three.
Answer:
The linear velocity of the object is 8.71 m/s.
Explanation:
Given;
mass of the object, m = 1 kg
radius of the circle, r = 3.3 meters
centripetal force, F = 23 N
Centripetal force is given by;

where;
v is the linear velocity of the object

Therefore, the linear velocity of the object is 8.71 m/s.
This question is incomplete, the missing image is uploaded along this answer.
Answer:
the coefficient of friction is 0.32
Explanation:
Given the data in the question;
we make use of kinematic equation of motion;
ω = ω₀ + ∝t
we substitute
ω = ( 0 rad/s ) + ( 0.4 rad/s² )( 9.903 s )
ω = 3.9612 rad/s
The centripetal force acting on the sample is;
Fc = mrω²
from the image; r = 200 mm = 0.2 m
so we substitute
Fc = m(0.2 m ) ( 3.9612 rad/s )²
Fc = (3.13822 m/s²)m
we know that the frictional force between the two materials should be providing the necessary centripetal force to rotate the sample object;
f = Fc
μN = Fc
μmg = (3.13822 m/s²)m
μ = (3.13822 m/s²)m / mg
μ = (3.13822 m/s²) / g
acceleration due to gravity g = 9.8 m/s²
so
μ = (3.13822 m/s²) / 9.8 m/s²
μ = 0.32
Therefore, the coefficient of friction is 0.32
Answer:
7.5 N/m
Explanation:
Potential energy of a spring can be calculated using below formula
Potential energy= 1/2kx^2
potential energy = 60 J
X= displacement = 4 m
K= spring constant=?
Substitute the values we have
60= 1/2 × K × 4^2
60= 1/2 × K × 16
60= K × 8
K= 7.5 N/m
Hence, the spring constant of the spring is 7.5 N/m