Answer:
0.2687 approximately 0.27
Explanation:
Diameter = 0.320
Speed = 40.0 rev/min
We are required to find coefficient of static friction between friction and button
The radius can be calculated as
0.320/2
= 0.160m
Then we have the rotational speed w = 40rev/min x 2pi/60
= 4.19 rad/s
umg = mrw²
u = mrw²/mg
u = rw²/g -------(1)
g = 9.8
When we put values into equation 1
0.150m x 4.19² / 9.8
= 0.150m x 17.5561 /9.8
= 0.2689
This is approximately 0.27
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
Between 1589-1592 when he discovered projecctile motion
Answer:
D) The acidic tomato juice reacted unfavorably with the aluminum pan
Answer:the first one was x
the second one is y
Explanation: