Ask question

Ask question

All categories

3 years ago

12

An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the

second harmonic of this pipe

1 answer:

bazaltina [42]3 years ago

5 0

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N + N--->N + N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

$L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda$

The frequency is calculated as follows;

$F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz$

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

You might be interested in

Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e

NeTakaya

We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

$Q=\frac{E}{Nr^2}$

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the <em>magnitude </em>of charge moved, Q is mathematically given as

$Q=\frac{E}{Nr^2}$

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

$Q=\frac{E}{Nr^2}$

For more information on this visit

brainly.com/question/16517842

3 0

2 years ago

Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto

Irina-Kira [14]

Answer:

W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

V = A L

V = π (R_ext ² - R_int ²) L

let's calculate

V = π (4² - 2²) 10⁻⁴ 3

V = 1.13 10⁻² m³

m = ρ V

m = 8966 1.13 10⁻²

m = 1.01 10² kg

the weight of the tube

W = mg

W = 1.01 10² 9.8

W = 9.93 10² N

4 0

3 years ago

Question 6 (5 points)

Ahat [919]

Answer:

its how life meets water and earth meets air.

Explanation:

4 0

3 years ago

Sphere A of mass 0.600 kg is initially moving to the right at 4.00 m/s. sphere B, of mass 1.80 kg is initially to the right of s

anzhelika [568]

A) The velocity of sphere A after the collision is 1.00 m/s to the right

B) The collision is elastic

C) The velocity of sphere C is 2.68 m/s at a direction of $-5.2^{\circ}$

D) The impulse exerted on C is 4.29 kg m/s at a direction of $-5.2^{\circ}$

E) The collision is inelastic

F) The velocity of the center of mass of the system is 4.00 m/s to the right

Explanation:

A)

We can solve this part by using the principle of conservation of momentum. The total momentum of the system must be conserved before and after the collision:

$p_i = p_f\\m_A u_A + m_B u_B = m_A v_A + m_B v_B$

$m_A = 0.600 kg$ is the mass of sphere A

$u_A = 4.00 m/s$ is the initial velocity of the sphere A (taking the right as positive direction)

$v_A$ is the final velocity of sphere A

$m_B = 1.80 kg$ is the mass of sphere B

$u_B = 2.00 m/s$ is the initial velocity of the sphere B

$v_B = 3.00 m/s$ is the final velocity of the sphere B

Solving for vA:

$v_A = \frac{m_A u_A + m_B u_B - m_B v_B}{m_A}=\frac{(0.600)(4.00)+(1.80)(2.00)-(1.80)(3.00)}{0.600}=1.00 m/s$

The sign is positive, so the direction is to the right.

B)

To verify if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

Before the collision:

$K_i = \frac{1}{2}m_A u_A^2 + \frac{1}{2}m_B u_B^2 =\frac{1}{2}(0.600)(4.00)^2 + \frac{1}{2}(1.80)(2.00)^2=8.4 J$

After the collision:

$K_f = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(0.600)(1.00)^2 + \frac{1}{2}(1.80)(3.00)^2=8.4 J$

The total kinetic energy is conserved: therefore, the collision is elastic.

C)

Now we analyze the collision between sphere B and C. Again, we apply the law of conservation of momentum, but in two dimensions: so, the total momentum must be conserved both on the x- and on the y- direction.

Taking the initial direction of sphere B as positive x-direction, the total momentum before the collision along the x-axis is:

$p_x = m_B v_B = (1.80)(3.00)=5.40 kg m/s$

While the total momentum along the y-axis is zero:

$p_y = 0$

We can now write the equations of conservation of momentum along the two directions as follows:

$p_x = p'_{Bx} + p'_{Cx}\\0 = p'_{By} + p'_{Cy}$ (1)

We also know the components of the momentum of B after the collision:

$p'_{Bx}=(1.20)(cos 19)=1.13 kg m/s\\p'_{By}=(1.20)(sin 19)=0.39 kg m/s$

So substituting into (1), we find the components of the momentum of C after the collision:

$p'_{Cx}=p_B - p'_{Bx}=5.40 - 1.13=4.27 kg m/s\\p'_{Cy}=p_C - p'_{Cy}=0-0.39 = -0.39 kg m/s$

So the magnitude of the momentum of C is

$p'_C = \sqrt{p_{Cx}^2+p_{Cy}^2}=\sqrt{4.27^2+(-0.39)^2}=4.29 kg m/s$

Dividing by the mass of C (1.60 kg), we find the magnitude of the velocity:

$v_c = \frac{p_C}{m_C}=\frac{4.29}{1.60}=2.68 m/s$

And the direction is

$\theta=tan^{-1}(\frac{p_y}{p_x})=tan^{-1}(\frac{-0.39}{4.27})=-5.2^{\circ}$

D)

The impulse imparted by B to C is equal to the change in momentum of C.

The initial momentum of C is zero, since it was at rest:

$p_C = 0$

While the final momentum is:

$p'_C = 4.29 kg m/s$

So the magnitude of the impulse exerted on C is

$I=p'_C - p_C = 4.29 - 0 = 4.29 kg m/s$

And the direction is the angle between the direction of the final momentum and the direction of the initial momentum: since the initial momentum is zero, the angle is simply equal to the angle of the final momentum, therefore $-5.2^{\circ}$ .

E)

To check if the collision is elastic, we have to check if the total kinetic energy is conserved or not.

The total kinetic energy before the collision is just the kinetic energy of B, since C was at rest:

$K_i = \frac{1}{2}m_B u_B^2 = \frac{1}{2}(1.80)(3.00)^2=8.1 J$

The total kinetic energy after the collision is the sum of the kinetic energies of B and C:

$K_f = \frac{1}{2}m_B v_B^2 + \frac{1}{2}m_C v_C^2 = \frac{1}{2}(1.80)(1.20)^2 + \frac{1}{2}(1.60)(2.68)^2=7.0 J$

Since the total kinetic energy is not conserved, the collision is inelastic.

F)

Here we notice that the system is isolated: so there are no external forces acting on the system, and this means the system has no acceleration, according to Newton's second law:

$F=Ma$

Since F = 0, then a = 0, and so the center of mass of the system moves at constant velocity.

Therefore, the centre of mass after the 2nd collision must be equal to the velocity of the centre of mass before the 1st collision: which is the velocity of the sphere A before the 1st collision (because the other 2 spheres were at rest), so it is simply 4.00 m/s to the right.

Learn more about momentum and collisions:

brainly.com/question/6439920

brainly.com/question/2990238

brainly.com/question/7973509

brainly.com/question/6573742

#LearnwithBrainly

8 0

4 years ago

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature (37Â° C) by warmi

Gelneren [198K]

Answer:

7.72 Liters

Explanation:

normal body temperature = T_body =37° C

temperature of ice water = T_ice =0°c

specfic heat of water = c_{water} =4186J/kg.°C

if the person drink 1 liter of cold water mass of water is = m = 1.0kg

heat lost by body is Qwater =mc_{water} ΔT

= mc{water} ( T_ice - T_body)

= 1.0×4186× (0 -37)

= -154.882 ×10^3 J

here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories

= 286×4186J

so number of liters of ice water have to drink is

n×Q_{water} =Q_{walk} n= Q_{walk}/ Q_{water}

= 286×4186J/154.882×10^3 J

= 7.72 Liters

7 0

3 years ago

Read 2 more answers