The net displacement at a point on the string where the pulses cross is 0.2 m.
The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.
A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.
At the post, the pulses are reflected and return along the string without losing any of their amplitude.
Now, let's say the ends are free.
There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.
Now, since A = 0.1 m
Then, 2A = 2(0.1) = 0.2 m
As a result, the net displacement at the string's intersection of two pulses is 0.2 m.
The correct option is (c).
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How much work in J does the string do on the boy if the boy stands still?
<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>
<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>
<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>
<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
Answer:
Frequency = 3.19 * 10^14 Hz or 1/s
Explanation:
Relationship b/w frequency and wavelength can be expressed as:
C = wavelength * frequency, where c is speed of light in vacuum which is 3.0*10^8 m/s.
Now simply input value (but before that convert wavelength into meters to match the units, you do this by multiply it by 10^-9 so it will be 940*10^-9)
3.0 * 10^8 = Frequency * 940 x 10^-9
Frequency = 3.19 * 10^14 Hz or 1/s
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 