Answer:
Em₀ = U = m g h
, Em_{f} = K = ½ m v²
Explanation:
When a car is on a ramp it has a certain amount of mechanical energy. At the highest point of the ramp the mechanical energy is fully potential given by
Em₀ = U = m g h
As part of this energy descends down the ramp, part of this energy is transformed into kinetic energy and has one part of each, even though the sum remains the initial energy
Em = K + U = ½ m v² + mg y
y <h
when it reaches the bottom of the ramp it has no height therefore there is no potential energy, all of it has been transformed into kinetic energy
Em_{f} = K = ½ m v²
This energy transformation is in the case that the friction force is zero.
If there is a friction force, it performs work against the low car, it is reflected in an increase in the internal energy (temperature) of the car. In this case the energy in the lower part is less than the initial one by a factor
= - fr L
therefore the numeraire values of the velocity are lower, due to the energy lost by friction.
Answer:c>a>b
Explanation:
Given
All the engine extracts same amount of heat(Q) from High-temperature reservoir
For a) 400 and 500 K
For b)500 K and 600K
For c) 400 K and 600 K
So c will give the highest amount of work
c>a>b
Answer:
Newton's Third Law of Motion
Explanation:
it explains that forces always come in action-reaction pairs. The Third Law states that for every action force, there is an equal and opposite reaction force
Choice-A is the correct one. It doesn't say it, but it means he'll see the ball reach the catcher's mitt first BEFORE HE HEARS IT slap the mitt.
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
Now At xp we have:
Which is a second order equation, using the quadratic formula to solve for xp would give us:
or
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that
BOTH are correct. This is simply explained by considring the following.
Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.
