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Lera25 [3.4K]
3 years ago
8

If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 wil

l not be "picked up" by this pickup?
Physics
1 answer:
Lina20 [59]3 years ago
4 0

Answer:

The answer to this question can be defined as follows:

Explanation:

Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.

Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.

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A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0
algol13

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats m_1 and m_2,  and the length of the rod L.

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$,

solving for v, we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for m_1, m_2, g, and L we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}} $

\boxed{ v= 1.64m/s}

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

7 0
3 years ago
Help!!
Roman55 [17]
D. Circulatory
The digestive system works very closely with the circulatory system to get the absorbed nutrients distributed through your body. The circulatory system also carries chemical signals from your endocrine system that control the speed of digestion.
5 0
3 years ago
A topographic map would best provide information about which area? O state boundaries O interstate highways O routes of minor ro
VladimirAG [237]

Answer:

I believe it is D

Explanation:

7 0
3 years ago
Read 2 more answers
What is the momentum of a man of mass 10kg when he walks with a uniform velocity of 2m/s.
Stolb23 [73]

Answer:

Momentum(p) = 20kgm/s

Explanation:

Given

Mass = 10kg

Velocity = 2m/s

Required

Calculate the momentum of the man

Momentum is calculated as thus

Momentum(p) = Mass(m) * Velocity(v)

or

p = mv

So; to solve this question; we simply substitute 10kg for mass and 2m/s for velocity in the above formula;

The formula becomes

Momentum(p) = 10kg * 2m/s

Momentum(p) = 10 * 2 * kg * m/s

Momentum(p) = 20kgm/s

Hence, the momentum of the man is 20kgm/s

5 0
3 years ago
A particularly beautiful note reaching your ear from a rare stradivarius violin has a wavelength of 39.1 cm. the room is slightl
raketka [301]
The wavelength of the note is \lambda = 39.1 cm = 0.391 m. Since the speed of the wave is the speed of sound, c=344 m/s, the frequency of the note is
f= \frac{c}{\lambda}=879.8 Hz

Then, we know that the frequency of a vibrating string is related to the tension T of the string and its length L by
f= \frac{1}{2L} \sqrt{ \frac{T}{\mu} }
where \mu=0.550 g/m = 0.550 \cdot 10^{-3} kg/m is the linear mass density of our string.
Using the value of the tension, T=160 N, and the frequency we just found, we can calculate the length of the string, L:
L= \frac{1}{2f}  \sqrt{ \frac{T}{\mu} } =0.31 m
8 0
3 years ago
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