Answer:
the distance in meters traveled by a point outside the rim is 157.1 m
Explanation:
Given;
radius of the disk, r = 50 cm = 0.5 m
angular speed of the disk, ω = 100 rpm
time of motion, t = 30 s
The distance in meters traveled by a point outside the rim is calculated as follows;

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m
Answer:
a) θ = 2500 radians
b) α = 200 rad/s²
Explanation:
Using equations of motion,
θ = (w - w₀)t/2
θ = angle turned through = ?
w = final angular velocity = 1420 rad/s
w₀ = initial angular velocity = 420
t = time taken = 5s
θ = (1420 - 420) × 5/2 = 2500 rads
Again,
w = w₀ + αt
α = angular accelaration = ?
1420 = 420 + 5α
α = 1000/5 = 200 rad/s²
Acceleration=velocity/time
acceleration=28/4.22
therefore, acceleration=6.64
Answer:
h = height of the hotel room from the ground floor = 237.4m
Explanation:
Change in Potential Energy of tourist = ΔPE = PE2 – PE1 = mgh
PE1 is the potential energy of tourist at the ground floor
PE1 is the potential energy of tourist at the top (hotel room)
Given
PE1 = − 2.01 × 10⁵ J
PE2 = 0J
PE2 – PE1 = mgh
0 – (− 2.01 × 10⁵ J) = mgh
2.01 × 10⁵ J = 86.4×9.8×h
h = 2.01 × 10⁵/(86.4×9.8) = 237.4m
<span>Kinetic energy increases and potential energy decreases.
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