Which light is most sensitive to the eyes?
1 answer:
You might be interested in
The coefficient of static friction between the chair and the floor is 0.67
Explanation:
Given:
Weight of the chair = 25kg
Force = 165 N (F_applied)
Force = 127 N (F_max)
To find: Coefficient of static friction
The “coefficient of static friction” between a chair and the floor is defined as the ration of maximum force to the normal force acting on the chair
μ_s=
The F_n is equal to the weight multiplied by its gravity
∴=mg
Thus the coefficient of static friction changes as
μ_s=
μ_{s} =
= 0.67
Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
- Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:
- Now, we need first to find the value of the angular acceleration, that we can get from the following expression:
- Since the machine starts from rest, ω₀ = 0.
- We know the value of ωf₁ (the operating speed) in rev/min.
- Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:
- Replacing by the givens in (2):
- Solving for α:
- Replacing (5) and Δt in (1), we get:
- in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:
- In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:
- First of all, we need to find the value of the angular acceleration during the second period.
- We can use again (2) replacing by the givens:
- ωf =0 (the machine finally comes to an stop)
- ω₀ = ωf₁ = 57.5 rev/sec
- Δt = 32 s
- Solving for α in (9), we get:
- Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:
- In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows: