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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
Answer:
a. Quadruped arm and opposite leg raise
Explanation:
Quadruped arm and opposite leg lift
- Kneel on the floor, lean forward and place your hands down.
- Keep your knees in line with your hips and hands directly under your shoulders.
- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.
- Go back to the starting position.
This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.
It is also used together with other exercises for the treatment of hyperlordosis.
Answer:
22 km/h
Explanation:
Given that,
Speed of Xavier, v = 14 km/h
He tosses a set of keys forward on the ground at 8 km/h, v' = 8 km/h
We need to find the speed of the keys relative to the ground. Let it is V.
As both Xavier and the keys are moving in same diretion. The relative speed wrt ground is given by :
V = v+v'
V= 14 + 8
V = 22 km/h
So, the speed of the keys relative to the ground is 22 km/h.
