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3 years ago

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A stone is thrown horizontally with an initial speed of 9 m/s from the edge of a cliff. A stop watch measures the stone's trajec

tory time from the top of the cliff to the bottom to be 4.7 s. What is the height of the cliff? Round to the nearest tenth of a meter

1 answer:

Nady [450]3 years ago

6 0

Answer:

the height of the cliff is equal to 108.241 m

Explanation:

given,

initial speed of the stone horizontally = 9 m/s

vertical speed of the stone = 0 m/s

time taken of the trajectory = 4.7 s

using equation of motion

$s = ut + \dfrac{1}{2}at^2$

$s =0 \times 4.7 + \dfrac{1}{2} \times 9.8 \times 4.7^2$

$s = 0 + 4.9 \times 4.7^2$

$s = 4.9 \times 22.09$

s = 108.241 m

hence, the height of the cliff is equal to 108.241 m

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Explanation:

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$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

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In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

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$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

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2 years ago

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1 year ago

Read 2 more answers

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Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

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a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

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At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

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$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

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$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

= -0.24J

Kinetic energy =

$\frac{1}{2} * 4Vf^2$

= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

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= 3.33 m/s

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