Because the coefficient of friction depends on the surface
Given:
Uniform distributed load with an intensity of W = 50 kN / m on an overhang beam.
We need to determine the maximum shear stress developed in the beam:
τ = F/A
Assuming the area of the beam is 100 m^2 with a length of 10 m.
τ = F/A
τ = W/l
τ = 50kN/m / 10 m
τ = 5kN/m^2
τ = 5000 N/ m^2<span />
Explanation:
To find the average of these numbers, we just have to add the three numbers together and divide by 3.
- 2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.38
- 1.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/3
- 0.95 + 1.61 + 0.56 = 3.12 / 3 = 1.04
- 0.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26
All the bats living in a cave form a __colony__of bats in the cave ecosystem.
Answer:
29.2 ft/s
Explanation:
The distance of the light's projection on the wall
y = 13 tan θ
where θ is the light's angle from perpendicular to the wall.
The light completes one rotation every 3 seconds, that is, 2π in 3 seconds,
Angular speed = w = (2π/3)
w = (θ/t)
θ = wt = (2πt/3)
(dθ/dt) = (2π/3)
y = 13 tan θ
(dy/dt) = 13 sec² θ (dθ/dt)
(dy/dt) = 13 sec² θ (2π/3)
(dy/dt) = (26π/3) sec² θ
when θ = 15°
(dy/dt) = (26π/3) sec² (15°)
(dy/dt) = 29.2 ft/s
