Answer:

Explanation:
Since the hoop is rolling on the floor so its total kinetic energy is given as

now for pure rolling condition we will have

also we have

now we will have


now by work energy theorem we can say



now solve for final speed

Answer:
638 m.
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 94 m/s
Final velocity (v) = 22 m/s
Time (t) = 11 s
Distance (s) =?
We can obtain the distance travelled by using the following formula:
s = (u + v) t /2
s = (94 + 22) × 11 /2
s = 116 × 11 /2
s = 1276 /2
s = 638 m
Thus, the distance travelled is 638 m.
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
The outer surface of the glass expands faster than the inside surface since glass is a fairly good thermal insulator. The outer surface tries to expand without the inner surface doing so at the same time. This is the secret of Pyrex glass. It is more conductive to heat and the inner and outer surface expand (and contract) at the same time.