Answer:
(A) We are using them faster than they are replenished by nature
Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them. Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.
Efficiency is calculated through dividing the actual mechanical advantage by the hypothetical mechanical advantage:
- the actual mechanical advantage is 9J because that's how much work the light bulb doing
- the hypo. mechanical advantage is 100J. Ideally, in a perfect world, the light bulb can convert 100J input into 100J output, but do to resistance and other factors it is not possible.

change the decimal to a percentage:

the light bulb had 9% efficiency
Answer:
The constant angular acceleration of the centrifuge = -252.84 rad/s²
Explanation:
We will be using the equations of motion for this calculation.
Although, the parameters of this equation of motion will be composed of the angular form of the normal parameters.
First of, we write the given parameters.
w₀ = initial angular velocity = 2πf₀
f₀ = 3650 rev/min = (3650/60) rev/s = 60.83 rev/s
w₀ = 2πf₀ = 2π × 60.83 = 382.38 rad/s
θ = 46 revs = 46 × 2π = 289.14 rad
w = final angular velocity = 0 rad/s (since the centrifuge come rest at the end)
α = ?
Just like v² = u² + 2ay
w² = w₀² + 2αθ
0 = 382.38² + [2α × (289.14)]
578.29α = -146,214.4644
α = (-146,214.4644/578.29)
α = - 252.84 rad/s²
Hope this Helps!!!
The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316