Answer:0.3
Explanation:
Given
velocity of car=15 m/s
truck brought to halt in a distance of 38 m
We know

Final velocity (v)=0


(deceleration)
Therefore minimum coefficient of friction \mu will be


Answer:
w = 25.05 rad / s
, α = 0.7807 rad / s²
, θ = 1972.75
Explanation:
This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off
θ = w₀ t - ½ α t²
α = (w₀t - θ) 2/t²
let's reduce the magnitudes to the SI system
w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s
θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad
let's calculate the angular acceleration
α = (55.5 39 - 1570.8) 2/39²
α = 0.7807 rad / s²
having the acceleration we can calculate the final speed
w = w₀ - ∝ t
w = 55.5 - 0.7807 39
w = 25.05 rad / s
the time to stop w = 0
0 = wo - alpha t
t = wo / alpha
t = 55.5 / 0.7807
t = 71.09 s
the angle traveled
w² = w₀⁹ - 2 α θ
w = 0
θ = w₀² / 2α
let's calculate
θ = 55.5 2 / (2 0.7807)
θ = 1972.75
Refer to the diagram shown below.
Let x = distance of the base of the ladder from the base of the hill.
By definition,
sin(75°) = x/15
x = 75*cos(75°) = 3.882 ft
Answer:
The distance between the base of the hill and the base of the ladder is 3.9 ft (nearest tenth)
Yes. If an object would be vertical, it would most likely have potential energy due to gravity.