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attashe74 [19]
3 years ago
13

An organ pipe is open at both ends. It is producing sound

Physics
1 answer:
ozzi3 years ago
3 0

To solve this problem we will apply the concepts related to the wavelength of its third harmonic.

It describes that the wavelength is equivalent to

\lambda = \frac{2}{3}L

Here,

\lambda = Wavelength

The wavelength is in turn described as a function that depends on the change of the speed as a function of the frequency, that is to say

\lambda = \frac{v}{f}

In this case the speed is equivalent to the speed of sound and the frequency was previously given, therefore

\lambda = \frac{343}{262}

\lambda = 1.3091m

Finally the length of the pipe would be

L= \frac{3}{2}(1.3091)

L = 1.963m

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An object with smaller mass and an object with a larger mass have the same kinetic energy. Which object has the higher momentum?
KatRina [158]

Answer:

object with larger mass

Explanation:

the momentum of an object is directly proportional to its mass. larger mass = more momentum

4 0
3 years ago
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What is the direction of the force for a negative charge moving downward in a magnetic field pointing to the left? out of the sc
AveGali [126]
It would be towards your side. Right hand thumb rule
4 0
3 years ago
The voltage V in a circuit that satisfies the law V = IR is slowly dropping as the battery wears out. At the same time, the resi
Anastasy [175]

Answer:

-0.5\times 10^{-4} A/s

Explanation:

We are given that

\frac{dV}{dt}=-0.01 V/s

R=600 ohms

I=0.04 A

\frac{dR}{dt}=0.5ohm/s

V=IR

\frac{dV}{dt}=\frac{\partial V}{dI}\frac{dI}{dt}+\frac{\partial V}{dR}\frac{dR}{dt}

\frac{dV}{dt}=R\frac{dI}{dt}+I\frac{dR}{dt}

Substitute the values

-0.01=600\times \frac{dI}{dt}+0.04\times 0.5

-0.01-0.04\times 0.5=600\frac{dI}{dt}

-0.03=600\frac{dI}{dt}

\frac{dI}{dt}=\frac{-0.03}{600}=-0.5\times 10^{-4}A/s

5 0
4 years ago
The middle-C hammer of a piano hits two strings, producing beats of 1.60 Hz. One of the strings is turned to 260.00 Hz. What fre
expeople1 [14]

Answer:

The frequencies the other string could have are 258.4 Hz and 261.6 Hz.

Explanation:

Given;

beat frequency, Fb = 1.60 Hz

frequency of the first string, F₁ = 260.00 Hz

frequency of the second string, F₂ = ?

Beat frequency is given as;

Fb = F₂ - F₁     or      Fb = F₁ - F₂

Fb + F₁ = F₂    or      F₂ = F₁ - Fb

1.6 + 260 = F₂   or   F₂ = 260 - 1.6

261.6 Hz = F₂  or   F₂ = 258.4 Hz

Therefore, the frequencies the other string could have are 258.4 Hz and 261.6 Hz.

3 0
3 years ago
A negative point charge q1 = 25 nC is located on the y axis at y = 0 and a positive point charge q2 = 10 nC is located at y =14
sergey [27]

Answer:

 y = 0.1 m

Explanation:

The electrical power for point loads is

         V = k \sum \frac{q_i}{r_i}k Sum qi / ri

in this case

         V = k (- \frac{q_1}{r_1 } + \frac{q_2}{r_2})

indicate that V = 0

        \frac{q_1}{r_1} = \frac{q_2}{r_2}

        r₂ = \frac{q_2}{q_1} r_1

the distance r1 is

         r₁ = y -0

the distance r2

         r₂ = 0.14 -y

we substitute

       

        0.14 - y = \frac{10}{25}  y

          y ( \frac{10}{25} + 1) = 0.14

          y 1.4 = 0.14

          y = 0.14 / 1.4

          y = 0.1 m

7 0
3 years ago
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