Ask question

Ask question

All categories

svet-max [94.6K]

2 years ago

9

A slide whistle is an open-closed tube with an adjustable plunger that changes the length. You are playing the slide whistle and

pushing the plunger in at 8 cm/s when the tube is 15 cm long. What is the rate of change of the sound in Hz/s?

1 answer:

Serjik [45]2 years ago

4 0

Answer:

df / ft = -n 12 n= 1, 3, 5, ...

Explanation:

The speed of sound is

v = λ f

In a whistle that we approach by an open tube at one end and closed at the other, standing waves occur, which has a node in the closed part and a maximum in the open pate, whereby wavelength and the distance of the tube are related, the fundamental wave is

λ₁ = 4L

The harmonics are

λ₃ = 4L / 3

λ₅ = 4L / 5

The general formula

λₙ = 4L / n

with n = 1, 3, 5,…

We substitute and clear in the first equation

f = v n / 4L n = 1, 3, 5,…

Let's use derivatives to find the frequency change

df / dt = v n /4 dL⁻¹ / dt

d / dt (1/L) = - 1 / L² dL / dt

Where dL / dt = 8 cm / s

We replace

df / dt = - n v / L2 dL / dt

Let's calculate

df / dt = - n 340/152 8

df / ft = -n 12 n= 1, 3, 5, ...

You might be interested in

A(n) 71.1 kg astronaut becomes separated from the shuttle, while on a space walk. She finds herself 70.2 m away from the shuttle

denpristay [2]

Answer:

10.347 minutes.

Explanation:

According to F = ma, she exerts force on camera of the magnitude

F = 0.67Kg*12m/ $s^{2}$ = 8.04N, assuming it took her one second to accelerate camera to 12m/s, then by newtons third law, which says every action has equal and opposite reaction , the camera exerts the same amount of force on the astronaut which gives her acceleration of a = $\frac{8.04N}{70.2Kg} = 0.1130801680m/s^2$ .

and velocity of V = 0.1130801680m/s.

at this velocity , the astronaut has to cover the distance of 70.2 meters, it will take her 620.7985075s = 10.347 min to get to the shuttle (using S = vt).

4 0

2 years ago

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

2 years ago

A star orbiting a black hole in a clockwise direction at a radial distance of 1.0 × 106 km is acted upon by a counterclockwise f

snow_tiger [21]

Answer:

$2.5 \times 10^{11} N-m$ upwards

Explanation:

Torque is the vector cross product of the force and radial distance.

$\tau = rF sin \theta$

$\tau = (1.0\times 10^6 \times 10^3) m \times 250 N\times sin 90^o \\\Rightarrow \tau= 2.5 \times 10^{11} N-m$

The direction of the torque would be perpendicular to the direction of the force and radial distance. The direction of the force is counter-clockwise. The direction of the torque would be upwards.

4 0

2 years ago

A baggage handler throws a 15 kg suitcase along the floor of an airplane luggage compartment with a speed of 1.2 m/s. The suitca

Hatshy [7]

Answer:

0.0367

Explanation:

The loss in kinetic energy results into work done by friction.

Since kinetic energy is given by

KE=0.5mv^{2}

Work done by friction is given as

W= umgd

Where m is the mass of suitacase, v is velocity of the suitcase, g is acceleration due to gravity, d is perpendicular distance where force is applied and u is coefficient of kinetic friction.

Making u the subject of the formula then we deduce that

$u=\frac {v^{2}}{2gd}$

Substituting v with 1.2 m/s, d with 2m and taking g as 9.81 m/s2 then

$u=\frac {1.2^{2}}{2*9.81*2}=0.0366972477064\approx 0.0367$

Therefore, the coefficient of kinetic friction is approximately 0.0367

7 0

2 years ago

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

aivan3 [116]

A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>

4 0

2 years ago

Read 2 more answers