The lower the value of the coefficient of friction, the lower the resistance to sliding.
<u>Explanation:</u>
The coefficient of friction defines as directly proportionate with the resisting force, which is the frictional force. Hence, if there seems a decrease at coefficient of friction, then it is sure that the frictional force decreases.
We know that the frictional force on a body, is the product of coefficient of frictions and the normal forces acting on the body. Note that friction acts only, if a body is in contact, and it is of three types, static, kinetic and rolling.
B
A is extremly hot
and c is -330.07 degrees farenheit
Answer:
The equation of equilibrium at the top of the vertical circle is:
\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}
The speed experimented by the car is:
\frac{N}{m}+g=\frac{v^{2}}{R}
v = \sqrt{R\cdot (\frac{N}{m}+g) }
v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}
v\approx 9.302\,\frac{m}{s}
The equation of equilibrium at the bottom of the vertical circle is:
\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}
The normal force on the car when it is at the bottom of the track is:
N=m\cdot (\frac{v^{2}}{R}+g )
N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)
N=21.690\,N
