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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They are positive and remain inside the nucleus.
Answer:
s^ -1 ( or 1/sec)
Explanation:
Velocity is given in units of displacement / sec
like feet /sec or m/sec
so b would have units of s^-1
(or perhaps a more general term would be time^-1)
I’m pretty sure it’s c sorry if I’m wrong
<span>553 ohms
The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is
Xc = 1/(2*pi*f*C)
where
Xc = Reactance in ohms
pi = 3.1415926535.....
f = frequency in hertz.
C = capacitance in farads.
I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance.
The 120 rad/s frequency is better known as 60 Hz.
Substitute known values into the formula.
Xc = 1/(2*pi* 60 * 0.00000480)
Xc = 1/0.001809557
Xc = 552.6213302
Rounding to 3 significant figures gives 553 ohms.</span>
