Answer: 0.306
Explanation:
from the question we are given the following
mass of sled (m) = 50 kg
force (f) = 1.75 x 10^2 N = 175 N
distance (s) = 6 m
net work done on the sled = 1.50 x 10 ^2 N = 150 N
acceleration due to gravity (g) = 9.8 m/s^2
coefficient of friction = μ
lets first calculate the frictional force (ff)
ff = μ x m x g = μ x 50 x 9.8 = 490 μ
work done on the slide by the applied force (W1)= f x s = 175 x 6 = 1050 j
work done on the slide by frictional force (W2) = ff x s = 490 μ x 6 = 2940μ j
now the net work done is the work done by the frictional force subtracted from the work done by the applied force
net work done = W1 - W2
150 = 1050 - 2940μ
2940μ = 1050 - 150
μ = 900 / 2940
μ = 0.306
this is a direct current circuit.
Answer:
The moment of inertia will be increased by a factor of 1575
Explanation:
Let the moment of inertia of a wheel be given as;
where;
I is the moment of inertia of the wheel
m is mass of the wheel
r is radius of the wheel
when the mass is increased by a factor of 7 and the radius is increased by a factor of 15
![I_1 = \frac{1}{2}mr^2\\ \\I_2 = \frac{1}{2}(7m)(15r)^2\\\\I_2 = \frac{1}{2}(7m)(225r^2)\\\\I_2 = 7\times 225[\frac{1}{2}(mr^2)]\\\\I_2 = 1575[\frac{1}{2}(mr^2)]\\\\Recall, \frac{1}{2}(mr^2) = I_1 \\\\I_2 = 1575 I_1](https://tex.z-dn.net/?f=I_1%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmr%5E2%5C%5C%20%5C%5CI_2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%287m%29%2815r%29%5E2%5C%5C%5C%5CI_2%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%287m%29%28225r%5E2%29%5C%5C%5C%5CI_2%20%3D%207%5Ctimes%20225%5B%5Cfrac%7B1%7D%7B2%7D%28mr%5E2%29%5D%5C%5C%5C%5CI_2%20%3D%201575%5B%5Cfrac%7B1%7D%7B2%7D%28mr%5E2%29%5D%5C%5C%5C%5CRecall%2C%20%5Cfrac%7B1%7D%7B2%7D%28mr%5E2%29%20%3D%20I_1%20%5C%5C%5C%5CI_2%20%3D%201575%20I_1)
Thus, the moment of inertia will be increased by a factor of 1575
We need more information such as how high it was falling from
