Two resistors, R1 and R2, are connected
1 answer:
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Answer:
.
Explanation:
The frequency of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)
The wave in this question travels at a speed of . In other words, the wave would have traveled
in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be
How many wave cycles can fit into that ? The wavelength of this wave
gives the length of one wave cycle. Therefore:
.
That is: there are wave cycles in
of this wave.
On the other hand, Because that of this wave goes through that point in each second, that
wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be
Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:
.
The calculations above can be expressed with the formula:
,
where
represents the speed of this wave, and
represents the wavelength of this wave.
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:
And the distance traveled downwards is:
If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:
Replacing into the first equation:
Rationalizing:
Let's call v1 the final speed of the package dropped from a height H. Thus:
Let v2 be the final speed of the package dropped from a height 4H. Thus:
Taking out the square root of 4:
Dividing v2/v1 we can compare the final speeds:
Simplifying:
The final speed of the second package is twice as much as the final speed of the first package.