Answer:
elastic potential energy
You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.
Explanation:
A. freezing, when water turns to ice the water is turning from a liquid to a solid.
Answer:
The answer is below
Explanation:
The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm
radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m
a) Initial angular velocity (
) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s
θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
angular acceleration (α) is:

b)
c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad
a) When it stops, the final angular velocity is 0. Hence:

θ = 323 rad
Answer:
1.25kg
Explanation:
Simply multiply volume and density together
A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
<h3>What is power?</h3>
In physics, power (P) is the work (W) done over a period of time.
- Step 1. Calculate the work done by the bodybuilder each time.
The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.
W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N
- Step 2. Calculate the work done by the bodybuilder over 10 times.
W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N
- Step 3. Calculate the power exerted by the bodybuilder.
The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.
P = 1.9 × 10⁴ N/45 s = 421 W
A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)
Learn more about power here: brainly.com/question/911620
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