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2 years ago

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A model rocket flies horizontally off the edge of a cliff at a velocity of 80.0m/s. If the canyon below is 128.0 m deep, how far

from the edge of the cliff does the model rocket land?
a. 112m
b. 225m
c. 337m
d. 409m

1 answer:

RSB [31]2 years ago

5 0

Answer:

c. 337

Explanation:

can someone answer my question

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A tray is moved horizontally back and forth in simple harmonic motion at a frequency of f = 2.07 Hz. On this tray is an empty cu

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Answer:

Explanation:

Given

Frequency of SHM is $f=2.07\ Hz$

Amplitude of SHM is $A=3.13\ cm$

Cup begins to slip when it overcomes the friction force

Friction force $F_s=\mu mg$

Applied force $F=ma$

$ma=\mu mg$

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and maximum acceleration during SHM is

$a=A\omega ^2$

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$a=3.13\times 10^{-2}\times (2\pi 2.07)^2$

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3 years ago

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Which of the following are true for acceleration?

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2 years ago

Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

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(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

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3 years ago