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3 years ago

15

If you blow across the open end of a soda bottle and produce a tone of 250 Hz, what will be the frequency of the next harmonic h

eard if you blow much harder?
___Hz

1 answer:

Naddika [18.5K]3 years ago

8 0

Answer:

Generally, the lowest overtone for a pipe open at one end and closed would be at y / 4 where y represents lambda, the wavelength.

Since F (frequency) = c / y Speed/wavelength

F2 / F1 = y1 / y2 because c is the same in both cases

F2 = y1/y2 * F1

F2 = 3 F1 = 750 /sec

Note that L = y1 / 4 = 3 y2 / 4 for these wavelengths to fit in the pipe

and y1 = 3 y2

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Answer:

$t=5.36\times 10^{-6}\ s$

Explanation:

Given that,

Lightning flashes one mile (1609 m) away from you.

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The answer is:

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3 years ago

Read 2 more answers

An ideal gas at 27°C is contained in a piston that ensures that its pressure will always be constant. Raising the tem- perature

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Answer:

T = 600K

Explanation:

See attachment below.

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4 years ago

A racquet ball with mass m = 0.256 kg is moving toward the wall at v = 11.8 m/s and at an angle of θ = 29° with respect to the h

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Answer:

Part a)

$P = 5.72 kg m/s$

Part b)

$\Delta P = 2.93 kg m/s$

Part c)

$F = 44.4 N$

Part d)

$\Delta P = 5.02 kg m/s$

Part e)

$\Delta t = 0.113 s$

Part f)

$\Delta K = 0$

Explanation:

As we know that initial velocity of the ball is given as

$v = 11.8 cos29 \hat i + 11.8 sin29 \hat j$

$v_i = 10.3 \hat i + 5.72 \hat j$

Now final velocity of the system is given as

$v_f = 10.3\hat i - 5.72\hat j$

Part a)

now magnitude of initial momentum is given as

$P = mv$

$P = 0.256(11.8)$

$P = 5.72 kg m/s$

Part b)

Change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(5.72 + 5.72)$

$\Delta P = 2.93 kg m/s$

Part c)

As we know that average force is defined as the rate of change in momentum

so here we have

$F = \frac{\Delta P}{\Delta t}$

$F = \frac{2.93}{0.066}$

$F = 44.4 N$

Part d)

Magnitude of change in momentum is given as

$\Delta P = m(v_f - v_i)$

$\Delta P = 0.256(7.8 + 11.8)$

$\Delta P = 5.02 kg m/s$

Part e)

As we know that in 2nd case the force is same as the initial force

so we will have

$\frac{\Delta P}{\Delta t} = F$

$\frac{5.02}{\Delta t} = 44.4$

$\Delta t = 0.113 s$

Part f)

Since this is elastic collision so change in kinetic energy must be ZERO

$\Delta K = 0$

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3 years ago