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-Dominant- [34]
2 years ago
13

The range of human hearing is roughly from twenty hertz to twenty kilohertz. Based on these limits and a value of 343 m/s for th

e speed of sound, what are the lengths of the longest and shortest pipes (open at both ends and producing sound at their fundamental frequencies) that you expect to find in a pipe organ?
Physics
1 answer:
omeli [17]2 years ago
4 0

Explanation:

It is given that, the range of human hearing is roughly from twenty hertz to twenty kilohertz.

Minimum frequency, f = 20 Hz

Maximum frequency, f' = 20,000 Hz

We need to find the lengths of the longest and shortest pipes. For open pipes, the length of pipe is given by :

l=\dfrac{v}{2f}

For shortest pipe, frequency should be maximum, l=\dfrac{343\ m/s}{2\times 20000\ Hz}

l = 0.008575 m

For longest pipe, frequency should be minimum, l'=\dfrac{343\ m/s}{2\times 20\ Hz}

l' = 8.575 m

So, the lengths of longest and shortest pipes are 8.575 meters and 0.008575 meters respectively.

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Density =mass/volume 20/10=2

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A shopper pushes a 5.32 kg grocery cart
Juli2301 [7.4K]

Answer:

\text { acceleration of the cart is } 10.94 \mathrm{m} / \mathrm{s}^{2}

Explanation:

According to “Newton's second law”

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Force = mass × acceleration

\text { Acceleration }=\frac{\text { force }}{\text { mass }}

Given that,

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\text { Force }=12.7 \mathrm{N} \text { forces at }-28.7^{\circ}

x=-28.7^{\circ}

F = 12.7N

Normal force = mg + F sinx,  

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“g” being the "acceleration of gravity",

“x” being the "angle of the cart"

\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}\text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^ 2 \text { on Earth })

To find normal force substitute the values in the formula,

Normal force = 5.32 × 9.8 + 12.7 × sin(-28.7)

Normal force = 52.136 + 12.7 × 0.480

Normal force = 52.136 + 6.096

Normal force = 58.232 N

<u>Acceleration of the cart</u>:

\text { Acceleration }=\frac{\text {Normal force}}{\text { mass }}

\text { Acceleration }=\frac{58.232}{5.32}

\text { Acceleration }=10.94 \mathrm{m} / \mathrm{s}^{2}

\text { Therefore, "acceleration of the cart" is } 10.94 \mathrm{m} / \mathrm{s}^{2}

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2 years ago
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Vi = Initial Velocity = 0 m/s (Since, ball starts from rest)

Therefore, using the values, we get:

2a(21.9 m) = (7.14 m/s)² - (0 m/s)²

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