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3 years ago

7

A yodeler yodels at a cliff 2000 m away. The sound of the echo returns in 9.8 s. If the wavelength of the sound wave is .85 m th

en what is the period of the sound wave?

1 answer:

stepladder [879]3 years ago

5 0

Answer:

Hey man, are you who I think you are? ... you used your student number on your account.... anyways im in your class :) guess who I am for free points. so the formula for period is Period=Wavelength/Velocity, so P=.85/velocity, to find velocity would be using the formula speed=distance/time, so the velocity is 2000/9.8 which is around 204.0816327.... so back to period formula, P=.85m/204.08 which gives a period of .004165. But im not entire sure so you have to check that on your own. (gl guessing who I am. hehe)

Explanation:

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An air-track glider attached to a spring oscillates between the 10 cm mark and the 60 cm mark on the track. The glider completes

Assoli18 [71]

Answer:

a) Time period is 3.3 seconds

b) The frequency is 0.3030 Hz

c) amplitude is 0.25 m

d) maximum speed is 0.476 m/s

Explanation:

Given the data in the question;

a) Period

Time Period T = Time taken for one oscillation

T = 33s / 10 = 3.3 seconds

Therefore, Time period is 3.3 seconds

b) Frequency

we know that frequency is the inverse of time period

so;

Frequency f = 1/T = 1 / 3.3 s

Frequency f = 0.3030 Hz

Therefore, The frequency is 0.3030 Hz

c) amplitude

amplitude A = $\frac{1}{2}$ ( 60 cm - 10 cm )

A = $\frac{1}{2}$ × 50 cm

A = 25 cm

A = 0.25 m

Therefore, amplitude is 0.25 m

d) maximum speed of the glider

maximum speed $V_{max}$ = ωA

and ω = 2π/T

so maximum speed $V_{max}$ = $\frac{2\pi }{T}$ A

so we substitute

so maximum speed $V_{max}$ = $\frac{2\pi }{3.3}$ × 0.25 m

so maximum speed $V_{max}$ = 0.476 m/s

Therefore, maximum speed is 0.476 m/s

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A mountain-climber friend with a mass of 74 kg ponders the idea of attaching a helium-filled balloon to himself to effectively r

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Answer:

$V=16.65 m^3$

Explanation:

The volume of the balloon can be find compared the force in each cases so:

reduce 25% from 74kg

$R=\frac{25}{100}*74kg=18.5kg$

So the net force uproad on the balloon is

$F_b=18.5kg*g$

Now the density of the both gases air and helium are different however the volume is the same change offcorss the mass so:

$P_h=\frac{m}{V}=0.179 kg/m^3$

$P_A=1.29 kg/m^3$

$F_b=F_A-F_H$

$F_b=m_a*g-m_h*g$

$m=P/V$

$18.5kg*g=(1.29kg/m^3-0.179kg/m^3*)V*g$

$V=\frac{18.5kg}{(1.29-0.179)kg/m^3}$

$V=16.65 m^3$

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The answer is C 100 km/h west

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4 years ago

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Answer:

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Explanation:

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3 years ago

Read 2 more answers

In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s

riadik2000 [5.3K]

Answer:

Part a)

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

$\tau = I\alpha$

here we will have

$\tau = (m_1g - m_2g)(\frac{L}{2})$

now we will have inertia of two masses given as

$I = (m_1 + m_2)(\frac{L}{2})^2$

now we have

$I = (m_1 + m_2)\frac{L^2}{4}$

now the angular acceleration is given as

$\alpha = \frac{\tau}{I}$

so we have

$\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}$

Part b)

Now if the rod is not massles then we will have total inertia given as

$I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}$

so we will have

$I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}$

now the acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}$

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