Explanation:
Given:
u = 20 m/s
a = 5 m/s^2
v = 30 m/s
t = ?
Use the first kinematic equation of motion:
v = u + at
t = (v - u)/a = 10/5 = 2 seconds
Answer:
Explanation:
We know that the frequency of the nth harmonic is given by 
, where 
is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:
Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):
Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):
So the tension is:
Which for our values is:
If it is the same vehicle, then the 60mph vehicle has more kinetic energy since it is moving faster. Therefore, it requires more energy to stop, and if it is the same car with the same beak system, the braking distance of the 30mph car will be significantly shorter than the 60mph car.
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.
