Answer:
Well it would be equal to 500N because pushing forward the ball (or whatever maybe a body) would push the canon back an even 500N backwards...
Explanation:
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2
Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.
-F=25(-1.8)
F=45N
Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.
45=u(25g)
45=u(25*10)
Therefore, the coefficient of friction is 0.18
Hope that helps
Answer:
0.76
Explanation:
we are given:
radius (r) =5.7 m
speed (s) = 1 revolution in 5.5 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
coefficient of friction (Uk) = ?
we can get the minimum coefficient of friction from the equation below
centrifugal force = frictional force
m x r x ω^{2} = Uk x m x g
r x ω^{2} = Uk x g
Uk = 
where ω (angular velocity) = 
= 
= 1.14
Uk = 
= 0.76
Answer:
R = 710.7N
L = 67.689 N
During gravity fall L = R = 0 N
Explanation:
So the acceleration that the elevator is acting on the woman (and the package) in order to result in a net acceleration of 0.15g is
g + 0.15g = 1.15g
The force R that the elevator exerts on her feet would be product of acceleration and total mass (Newton's 2nd law):
a(m + M) = 1.15g(57 + 6) = 1.15*9.81*63 = 710.7N
The force L that she exerts on the package would be:
am = 1.15g *6 = 1.15*9.81*6 = 67.689N
When the system is falling, all have a net acceleration of g. So the acceleration that the elevator exerts on the woman (and the package) is 0, and so are the forces L and R.
