200 joules of work energy are involved. That's all we need to know to answer the question. Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath. Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.
Power = (work or energy) / (time to do the work or move the energy)
Power = (200 joules) / (5 seconds)
<em>Power = 40 watts</em>
This is another time to look at Newton's 2nd law of motion:
Net Force = (mass) x (acceleration)
If the object is not moving, then its acceleration is certainly zero, and Newton's law looks like this:
Net Force = (mass) x (zero)
or Net Force = (zero) .
"Net Force = zero" means that if there ARE any forces acting on the object, then they add up to zero, and we call them "balanced" forces.
So the answer is '<em>yes</em>', and that's why.
Answer:
D. A Molecule
Explanation:
<em>The smallest particle of a compound that still has the properties of the compound is a molecule. It is composed of two or more bonded atoms.</em>
<span>If Shelly rolls ball A in the positive x direction with a velocity of 7.5 meters/second, and It hits stationary ball B and they undergo elastic collision, thus the two balls have different masses, then the following statement which is true is the statement that stated that there was no y-momentum initially.</span>
Answer:
the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
Explanation:
This problem can be solved using the kinematics relations, let's start by finding the final velocity of the acceleration period
v² = v₀² + 2 a₁ x
indicate that the initial velocity is zero
v² = 2 a₁ x
let's calculate
v = 
v = 143.666 m / s
now for the second interval let's find the distance it takes to stop
v₂² = v² - 2 a₂ x₂
in this part the final velocity is zero (v₂ = 0)
0 = v² - 2 a₂ x₂
x₂ = v² / 2a₂
let's calculate
x₂ = 
x₂ = 573 m
as the stopping distance is greater than the free length of the track, the vehicle leaves the track before it can brake
