Answer:
the vertical acceleration of the case is 1.46 m/s
Explanation:
Given;
mass of the clarinet case, m = 3.07 kg
upward force applied by the man, F = 25.60 N
Apply Newton's second law of motion;
the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration
F = mg + m(-a)
the acceleration is negative due to downward motion from the top of the piano.
F = mg - ma
ma = mg - F

Therefore, the vertical acceleration of the case is 1.46 m/s²
Answer with explanation :
The negative sign means that the potential energy decreases by the movement of the electron.
negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.
Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.
Answer: W = 294 J
Explanation: Solution:
Work is expressed as the product of force and the distance of the object.
W = Fd where F = mg
W= Fd
= mg d
= 15 kg ( 9.8 m/s²) ( 2m )
= 294 J
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ =
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ =
F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
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