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1 year ago

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A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or

gan pipe with one end closed and one end open. What is the tension in the string if the speed of sound in air is 344 m/s?

1 answer:

love history [14]1 year ago

6 0

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

$f=\dfrac{nV}{4L}$

$f=\dfrac{3\times 344}{4\times 2}$

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²) x 1.5 x 10⁻³ x 100

T=1022.42 N

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When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$

Solving the above quadratic equation

$\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}$

$\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega$

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

6 0

1 year ago

Two identical horizontal sheets of glass have a thin film of air of thickness t between them. The glass has refractive index 1.4

Gre4nikov [31]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= $(m+ \frac{1}{2} \lambda)$

replacing ;

Δx = 2t ; we have:

2t = $(m+ \frac{1}{2} \lambda)$

Given that thickness t = 700 nm

Then

2× 700 = $(m+ \frac{1}{2} \lambda)$ --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = $(m+ \frac{1}{2} \lambda)$ ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus; the wavelength λ of the light when it is traveling in air = 560 nm

b)

For the smallest thickness $t_{min} ; \ \ \ m =0$

∴ $2t_{min} =\frac{\lambda}{2}$

$t_{min} =\frac{\lambda}{4}$

$t_{min} =\frac{560}{4}$

$t_{min} =140 \ \ nm$

Thus, the smallest thickness t of the air film = 140 nm

7 0

2 years ago

Read 2 more answers

The reaction energy of a reaction is the amount of energy released by the reaction. It is found by determining the difference in

solmaris [256]

Answer: the correct answer is 7.8026035971 x 10^(-13) joule

Explanation:

Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get

m = 4.00260 u + 222.01757 u = 226.02017 u .

The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is

Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,

which is equivalent to an energy change of

Delta E = (0.00523 u)*(931.5MeV/1u)

Delta E = 4.87 MeV

Converting 4.87 MeV to Joules

1 joule [J] = 6241506363094 mega-electrón voltio [MeV]

4 mega-electrón voltio = 6.40870932 x 10^(-13) joule

4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule

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2 years ago

The total charge that an automobile battery can supply without being recharged is given in terms of ampere-hours. A typical 12 V

Lelechka [254]

Answer:

7.894 Hours.

Explanation:

Based on information number hours that this battery will last with give load has mathematical relation of.

$t = \frac{60Ah}{load in amperes.}$

with load 60A t = 1h, 30A t = 2h so on and forth.

two head lights draw total current of 2x3.8A = 7.6A.

putting this in above relation gives.

$t = \frac{60Ah}{7.6A}=7.894 h$ .

That is how long will it be before battery is dead.

6 0

1 year ago

A capacitor, C1, consists of two parallel circular plates with radius R and separation of d. A second capacitor, C2, consists of

qaws [65]

Answer:

$\frac{E_1}{E_2}= 4$

Explanation:

Capacitance C is given by

$C= \frac{\epsilon_0A}{d}$

A= area of capacitor cross section

d= distance

therefore,

$C_1= \frac{\epsilon_0A_1}{d_1}$

A_1= πR^2

d_1= d

$C_2= \frac{\epsilon_0A_2}{d_2}$

A_= π(2R)^2

d_2 = 2d

$q= \frac{\epsilon_0A_1}{d_1}V_1$

threfore

$V_1= \frac{qd_1}{\epsilon_0A_1}$

and

$V_2= \frac{qd_2}{\epsilon_0A_2}$

also we know that E= V/d

⇒ $\frac{E_1}{E_2}= \frac{V_1}{V_2}\times\frac{d_2}{d_1}$

⇒ $\frac{E_1}{E_2}= \frac{qd_1}{\epsilon_0A_1}\times\frac{\epsilon_0A_2}{qd_2}\times\frac{d_2}{d_1}$

= A_1/A_2= $\frac{4R^2}{R^2}$ =4

therefore,

$\frac{E_1}{E_2}= 4$

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1 year ago