Answer:
Explanation:
In wheel and axle. …with the system is the velocity ratio, or the ratio of the velocity (VF) with which the operator pulls the rope at F to the velocity at which the weight W is raised (VW). This ratio is equal to twice the radius of the large drum divided by the difference…
The longer you spend reading and thinking about this question,
the more defective it appears.
-- In each case, the amount of work done is determined by the strength
of
the force AND by the distance the skateboard rolls <em><u>while you're still
</u></em>
<em><u>applying the force</u>. </em>Without some more or different information, the total
distance the skateboard rolls may or may not tell how much work was done
to it.<em>
</em>
-- We know that the forces are equal, but we don't know anything about
how far each one rolled <em>while the force continued</em>. All we know is that
one force must have been removed.
-- If one skateboard moves a few feet and comes to a stop, then you
must have stopped pushing it at some time before it stopped, otherwise
it would have kept going.
-- How far did that one roll while you were still pushing it ?
-- Did you also stop pushing the other skateboard at some point, or
did you stick with that one?
-- Did each skateboard both roll the same distance while you continued pushing it ?
I don't think we know enough about the experimental set-up and methods
to decide which skateboard had more work done to it.
Answer:
The value of each charge is 4.22 x 10⁻⁵ C
Explanation:
Given;
distance between the two identical charges, d = 2 m
the force of repulsion between these two charges, F = 4N
Apply Coulomb's law;

Therefore, the value of each charge is 4.22 x 10⁻⁵ C
Here, "Wavelength is same for both waves" it is the distance between two crests or two consecutive troughs, so, it is constant for both of them, you can easily figure it out.
In short, Your Answer would be "Wavelength"
Hope this helps!
Answer:
The value of acceleration that accomplishes this is 8.61 ft/s² .
Explanation:
Given;
maximum distance to be traveled by the car when the brake is applied, d = 450 ft
initial velocity of the car, u = 60 mph = (1.467 x 60) = 88.02 ft/s
final velocity of the car when it stops, v = 0
Apply the following kinematic equation to solve for the deceleration of the car.
v² = u² + 2as
0 = 88.02² + (2 x 450)a
-900a = 7747.5204
a = -7747.5204 / 900
a = -8.61 ft/s²
|a| = 8.61 ft/s²
Therefore, the value of acceleration that accomplishes this is 8.61 ft/s² .