Answer:
Kinetic energy and potential energy.
The term 'mechanical energy' refers to the sum of the kinetic energy and the gravitational potential energy of an object,
Answer:
So A we cant sadly do because we cant draw. B is going to be kinetic. Thats because static friction means it stays in one place, for kinetic it means moving. So it will be 0.05 as the coefficient of the friction. Sadly, I cannot calculate C. You will have to use trigonemetry but I cannot fit that big an explanation.
Answer to A: the free body diagram would be the ski things inclined with gravity, friction, and air resistance. I except you know which directions
Answer to B: Kinetic friction is the answer.
Answer to C: Find on own, I cannot write super big explanations - use trigonometry.
Answer:
(a) The current should be in opposite direction
(b) The current needed is 39.8 A
Explanation:
Part (a)
Based, on right hand rule, the current should be in opposite direction
Part (b)
given;
strength of magnetic field, B = 370 µT
distance between the two parallel wires, d = 8.6 cm

At the center, the magnetic field strength is twice

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

Therefore, current needed is 39.8 A
Newton's 3 laws are...
inertia: things tend to continue to do what they are doing.
Change: to make something change you need a force to change it. the force needed = the mass times its acceleration
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Resistance: When you push on something, it pushes back.
From yahoo answers
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Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2 
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²