A sinusoidal transverse wave travels along a long, stretched string. The amplitude of this wave is 0.08190.0819 m, its frequency
1 answer:
Answer:
The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
Explanation:
Given that,
Amplitude = 0.08190 m
Frequency = 2.29 Hz
Wavelength = 1.87 m
(a). We need to calculate the shortest transverse distance between a maximum and a minimum of the wave
Using formula of distance
Where, d = distance
A = amplitude
Put the value into the formula
Hence, The shortest transverse distance between a maximum and a minimum of the wave is 0.1638 m.
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Answer:
(a) 25 m
(b) 75 m
Explanation:
Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.
So, the speed of the jogger,
Let d be the distance covered by him in time, t s.
As distance=(speed) x (time)
So,
From equation (i)
As the jogger starts from origin, so, the distance, , also represents the position of the jogger at the time
s.
The position-time graph has been shown.
(a) From equation (ii), for t=5.0 s
So, the jogger is at a distance of 25 m from the origin.
(b) Similarly, for t=15.0 s
So, the jogger is at a distance of 75 m from the origin.
