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svet-max [94.6K]

4 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]4 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, ...

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bazaltina [42]

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6 0

3 years ago

An unknown charged particle passes without deflection throughcrossed electric and magnetic fields of strengths 187,500 V/m and0.

UNO [17]

Explanation:

The given data is as follows.

Electric field strength (E) = 187,500 V/m

Magnetic field strength (B) = 0.125 T

Diameter (d) = 25.05 cm = 0.2505 m (as 1 m = 100 cm)

Radius (r) = $\frac{d}{2}$

= $\frac{0.2505}{2}$

= 0.12525 m

Formula to calculate the magnetic force ( $F_{M}$ ) is as follows.

$F_{M} = Bqv$ ............ (1)

Electrical force is calculated as follows.

$F_{E} = qE$ ............ (2)

On both electric and magnetic fields the velocity is perpendicular.

$F_{M} - F_{E}$ = 0

or, $F_{M} = F_{E}$

Hence, from equations (1) and (2)

Bqv = qE

or, v = $\frac{E}{B}$ ............. (3)

= $\frac{187500 V/m}{0.125 T}$

= 1,500,000 m/s

As the particle is moving in a semi-circular trajectory and motion of charged particle is given by the electric field as follows.

$F_{c} = \frac{mv^{2}}{r}$ ........... (4)

where, $F_{c}$ = centripetal force

$F_{M} = F_{c}$

Using equation (1) and (4) as follows.

$F_{M} = F_{c}$

Bqv = $\frac{mv^{2}}{r}$

$\frac{q}{m} = \frac{v}{Br}$

= $\frac{15 \times 10^{5}}{0.125 \times 0.12525}$

= $958.08 \times 10^{5} C/kg$

Thus, we can conclude that charge-to-mass ratio of the given particle is $958.08 \times 10^{5} C/kg$ .

8 0

3 years ago

A 1.0 kg copper rod rests on two horizontal rails 1.0 m apart and carries a current of 50 A from one rail to the other.

vagabundo [1.1K]

Answer

given,

mass of copper rod = 1 kg

horizontal rails = 1 m

Current (I) = 50 A

coefficient of static friction = 0.6

magnetic force acting on a current carrying wire is

F = B i L

Rod is not necessarily vertical

$F_x =i L B_d$

$F_y= i L B_w$

the normal reaction N = mg-F y

static friction f = μ_s (mg-F y )

horizontal acceleration is zero

$F_x-f = 0$

$iLBd = \mu_s(mg-F_y )$

B_w = B sinθ

B_d = B cosθ

iLB cosθ= μ_s (mg- iLB sinθ)

$B = \dfrac{\mu_smg}{i(cos\theta +\mu_s sin\theta)}$

$\theta =tan{-1}{\mu_s}$

$\theta =tan{-1}{0.6}$

$\theta = 31^0$

$B = \dfrac{0.6\times 1 \times 9.8}{50(cos31^0 +0.6 sin31^0)}$

B = 0.1 T

4 0

3 years ago

Identify the normal force on the shopping cart after 75 newtons of groceries are added to the cart

Anna71 [15]

If the shopping cart is parallel to the ground, the weight is acting vertically downward, then force is equal to weight. The normal force will be acting perpendicular to the ground,and acting upward opposite of the weight.

3 0

3 years ago

Read 2 more answers

Assume that a uniform magnetic field is directed into thispage. If an electron is released with an initial velocity directedfrom

Feliz [49]

Answer:

B). to the right

Explanation:

Since the direction of magnetic field is into the page

So here we know that

$B = B_o(-\hat k)$

now the velocity is from bottom to top

so we have

$v = v_o \hat j$

now the force on the moving charge is given as

$\vec F = q(\vec v \times \vec B)$

now we have

$\vec F = (-e)(v_o \hat j \times B_o(-\hat k))$

$\vec F = e v_o B \hat i$

so force will be towards Right

7 0

3 years ago