Answer: coefficient of static friction
= 0.31
Explanation: Since they negotiate the curve without skidding, the frictional force (F1) equals the centripetal force (F2).
F1= uN
F2 = M*(v²/r)
M is the combined mass 450kg
V is the velocity 18m/s
r is the radius 106m
N is the normal reaction 4410N
u is the coefficient of static friction
Making u subject of the formula we have that,
u = {450*(18²/106)} /4410
=1375.47/4410
=0.31
NOTE: coefficient of friction is dimensionless. It as no Unit.
Explanation:
The chemical properties of an element are determined by the configuration of its electrons in orbit around its nucleus. ... See a Periodic Table of the Elements. The number of protons in the nucleus of an atom is its Atomic Number.
For n resistors in series, the equivalent resistance is given by the sum of the resistances:
In this problem, we have three resistors, so the equivalent resistance of the load is the sum of the resistances of the three resistors:
It would result in stiffness and pain. Additional stress would be put onto the joints as well. Hope this helps, cheers!
Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
