If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate
2 answers:
Explanation:
It is given that,
The stone reaches the water surface is 3 seconds. Let d is the distance covered by the by the stone before hitting the water. Initial speed of the stone, u = 0
Using second equation of motion to find the distance covered.
Here, a = g
d = 44.1 meters
So, the stone will cover a distance of 44.1 meters before hitting the water. Hence, this is the required solution.
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Corrected and Formatted Question:
A wave on a string is described by y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))], where x is in m and t in s.
(a) In what direction is this wave traveling?
(b) What are the wave speed, frequency, and wavelength?
(c) At t = 0.50 , what is the displacement of the string at x = 0.20 m?
Answer:
The wave is travelling in the negative x direction
The wave speed = 12.0m/s
The frequency = 5Hz
The wavelength = 2.4m
The displacement at t = 0.50s and x = 0.20m is -0.029m
Explanation:
The general wave equation is given by;
y(x, t) = y cos (2(x/λ) - 2
ft) --------------------------------(i)
Where;
y(x, t) is the displacement of the wave at position x and a given time t
y = amplitude of the wave
f = frequency of the wave
λ = wavelength of the wave
Given;
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))] ------------------(ii)
Which can be re-written as;
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))] -------------(iii)
Comparing equations (i) and (iii) we have that;
=> 2π(x/(2.4 m) = 2π(x/λ)
=> λ = 2.4m
Therefore the wavelength of the wave is 2.4m
Also, still comparing the two equations;
=> 2π(t/(0.20 s) = 2πft
=> f = 1 / 0.20
=> f = 5Hz
Therefore the frequency of the wave is 5Hz
To get the wave speed (v), it is given by;
v = f x λ
Where f = 5Hz and λ = 2.4m
=> v = 5 x 2.4
=> v = 12.0m/s
Therefore, the speed of the wave is 12.0m/s
At t = 0.50s and x = 0.20m;
The displacement, y(x,t) of the string wave is given by
y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]
<em>Convert the amplitude of 3.0cm to m</em>
=> 3.0cm = 0.03m
<em>Substitute this back into the equation</em>
=> y(x, t) = (0.03m) × cos[2π(x/(2.4 m) + t/(0.20 s))]
<em>Substitute the values of t and x into the equation above;</em>
=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m) + 0.50/(0.20 s))]
<em>Carefully solve the equation</em>
=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]
=> y(x, t) = (0.03m) × cos[0.08π + 5π]
=> y(x, t) = (0.03m) × cos[5.08π]
=> y(x, t) = (0.03m) × cos[15.96]
=> y(x, t) = (0.03m) × cos[15.96]
=> y(x, t) = (0.03m) × -0.9684
=> y(x, t) = 0.029m
Therefore the displacement at those points is -0.029m
Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.