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An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of

Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

$3\times \frac{\lambda}{2} = 0.39$

$\lambda = 0.26 m$

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

$d = \frac{\lambda}{4}$

$d = \frac{0.26}{4} = 0.065 m = 6.5 cm$

So the distance will be 6.5 cm

3 0

3 years ago

3. A certain wire, 3 m long, stretches by 1.2 mm when under tension of 200 N. By how much does

nikitadnepr [17]

Answer:

The extension of the second wire is $e_2 = 0.0024 \ m = 2.4 mm$

Explanation:

From the question we are told that

The length of the wire is $L = 3 \ m$

The elongation of the wire is $e = 1.2mm = \frac{1.2}{1000} = 0.0012 m$

The tension is $F = 200 \ N$

The length of the second wire is $L_2 = 6 \ m$

Generally the Young's modulus(Y) of this material is

$Y = \frac{stress}{strain }$

Where $stress = \frac{F}{A}$

Where A is the area which is evaluated as

$A = \pi r^2$

and $strain = \frac{extention}{length} = \frac{e}{L}$

So

$Y = \frac{\frac{F}{\pi r^2 } }{ \frac{e}{L} }$

Since the wire are of the same material Young's modulus(Y) is constant

So we have

$\frac{F * L }{r^2 e} = \pi * Y = constant$

$F * L = constant * r^2 e$

Now the ration between the first and the second wire is

$\frac{F_1}{F_2} * \frac{L_1}{L_2} = \frac{r*2_1}{r^2} * \frac{e_1}{e_2}$

Since tension , radius are constant

We have

$\frac{L_1}{L_2} = \frac{e_1}{e_2}$

substituting values

$\frac{3}{6} = \frac{0.0012}{e_2}$

$0.5 e_2 = 0.0012$

$e_2 = \frac{ 0.0012 }{0.5}$

$e_2 = 0.0024 \ m = 2.4 mm$

3 0

4 years ago

In the absence of air resistance, a ball of mass m is tossed upward to reach a height of 20 m. at the 10 m position, halfway up,

Mazyrski [523]

On the ball, there is mg of net force.

<h3>How would one describe air resistance?</h3>

Air exerts a force known as air resistance. The force works in the opposite direction of an object traveling through the air.
While a sports vehicle with a streamlined design will encounter reduced air resistance and experience less drag, the automobile will be able to move more quickly than a truck with a flat front.

<h3>What Causes Air Resistance?</h3>

Air resistance, also referred to as "drag," is a force brought on by air. When air specks collide with an object's front, it slows down.
The more air particles that impact the object and the larger its surface area, the more resistance it faces.

learn more about air resistance here

brainly.com/question/1385438

#SPJ4

5 0

2 years ago

A mechanic turns a wrench using a force of 27.00 N at a distance of 16.00 cm from the rotation axis. The force is perpendicular

lutik1710 [3]

Answer:4.32Nm

Explanation:

The magnitude of the torque will be the product of the force and its perpendicular distance from the force.

Force = 27N

Perpendicular distance = 16cm = 0.16m

Torque = 27×0.16

Torque = 4.32Nm

3 0

3 years ago

An airplane traveling 919 m above the ocean at 505 m/s is to drop a case of twinkies to the victims below. How much time before

Elena L [17]

Answer:

a. 13.7 s b. 6913.5 m

Explanation:

a. How much time before being directly overhead should the box be dropped?

Since the box falls under gravity we use the equation

y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.

So,

y = ut - 1/2gt²

y = 0 × t - 1/2gt²

y = 0 - 1/2gt²

y = - 1/2gt²

t² = -2y/g

t = √(-2y/g)

So, t = √(-2 × 919 m/-9.8 m/s²)

t = √(-1838 m/-9.8 m/s²)

t = √(187.551 m²/s²)

t = 13.69 s

t ≅ 13.7 s

So, the box should be dropped 13.69 s before being directly overhead.

b. What is the horizontal distance between the plane and the victims when the box is dropped?

The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m

6 0

3 years ago