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2 answers:
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Answer:
(a). The height of the cliff is 41.67 m.
(b). The maximum height of the ball is 41.67 m
(c). The ball's impact speed is 16.52 m/s.
Explanation:
Given that,
Speed = 33 m/s
Angle = 60°
Time = 3.0 sec
(a). We need to calculate the height of the cliff
Using equation of motion
Put the value into the formula
(b). We need to calculate the maximum height of the ball
Using formula of height
Put the value into the formula
(c). We need to calculate the vertical component of velocity of ball
Using equation of motion
Put the value into the formula
We need to calculate the horizontal component of velocity of ball
Using formula of velocity
Put the value into the formula
We need to calculate the ball's impact speed
Using formula of velocity
Put the value into the formula
Hence, (a). The height of the cliff is 41.67 m.
(b). The maximum height of the ball is 41.67 m
(c). The ball's impact speed is 16.52 m/s.
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.