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charle [14.2K]
3 years ago
7

Learning Goal:

Physics
1 answer:
enot [183]3 years ago
6 0

Answer:

A. U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

Explanation:

The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

C=\dfrac{\epsilon A}{d}

C is the capacitance, A is the common plate area, d is the plate separation and \epsilon is the permittivity of the material between the plates.

For air or free space, \epsilon is \epsilon_0 called the permittivity of free space. In general, \epsilon=\epsilon_r \epsilon_0 where \epsilon_r is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum, \epsilon_r=1.

The energy stored in a capacitor is the average of the product of its charge and voltage.

U = \dfrac{QV}{2}

Its charge, Q, is related to its capacitance by Q=CV (this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for U,

U = \dfrac{CV^2}{2}

A. Substituting for C in U,

U_0 = \dfrac{\epsilon_0 A V^2}{2d}

B. When the distance is 3d,

U_1 = \dfrac{\epsilon_0 A V^2}{2\times3d}

U_1 = \dfrac{\epsilon_0 A V^2}{6d}

C. When the distance is restored but with a dielectric material of dielectric constant, K, inserted, we have

U_2 = \dfrac{K\epsilon_0 A V^2}{2d}

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A normal mode of a closed system is an oscillation of the system in which all parts oscillate at a single frequency. In general
valentina_108 [34]

Answer:

Explanation:

(A)

The string has set of normal modes and the string is oscillating in one of its modes.

The resonant frequencies of a physical object depend on its material, structure and boundary conditions.

The free motion described by the normal modes take place at the fixed frequencies and these frequencies is called resonant frequencies.

Given below are the incorrect options about the wave in the string.

• The wave is travelling in the +x direction

• The wave is travelling in the -x direction

• The wave will satisfy the given boundary conditions for any arbitrary wavelength \lambda_i

• The wave does not satisfy the boundary conditions y_i(0;t)=0


Here, the string of length L held fixed at both ends, located at x=0 and x=L

The key constraint with normal modes is that there are two spatial boundary conditions,y(0,1)=0


and y(L,t)=0

.The spring is fixed at its two ends.

The correct options about the wave in the string is

• The wavelength \lambda_i  can have only certain specific values if the boundary conditions are to be satisfied.

(B)

The key factors producing the normal mode is that there are two spatial boundary conditions, y_i(0;t)=0 and y_i(L;t)=0, that are satisfied only for particular value of \lambda_i  .

Given below are the incorrect options about the wave in the string.

•  A_i must be chosen so that the wave fits exactly o the string.

• Any one of  A_i or \lambda_i  or f_i  can be chosen to make the solution a normal mode.

Hence, the correct option is that the system can resonate at only certain resonance frequencies f_i and the wavelength \lambda_i  must be such that y_i(0;t) = y_i(L;t)=0


(C)

Expression for the wavelength of the various normal modes for a string is,

\lambda_n=\frac{2L}{n} (1)

When n=1 , this is the longest wavelength mode.

Substitute 1 for n in equation (1).

\lambda_n=\frac{2L}{1}\\\\2L

When n=2 , this is the second longest wavelength mode.

Substitute 2 for n in equation (1).

\lambda_n=\frac{2L}{2}\\\\L

When n=3, this is the third longest wavelength mode.

Substitute 3 for n in equation (1).

\lambda_n=\frac{2L}{3}

Therefore, the three longest wavelengths are 2L,L and \frac{2L}{3}.

(D)

Expression for the frequency of the various normal modes for a string is,

f_n=\frac{v}{\lambda_n}

For the case of frequency of the i^{th} normal mode the above equation becomes.

f_i=\frac{v}{\lambda_i}

Here, f_i is the frequency of the i^{th} normal mode, v is wave speed, and \lambda_i is the wavelength of i^{th} normal mode.

Therefore, the frequency of i^{th} normal mode is  f_i=\frac{v}{\lambda_i}

.

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Explanation:

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80 x 2.5 = 200 km/hr.

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3 years ago
Water moves through a constricted pipe in steady, ideal flow. At the
Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: 1.8\cdot 10^{-3} m^3/s

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2

where

p_1=1.75\cdot 10^4 Pa is the pressure in the lower section of the tube

h_1 = 0 is the heigth of the lower section

\rho=1000 kg/m^3 is the density of water

g=9.8 m/s^2 is the acceleration of gravity

v_1 is the speed of the water in the lower pipe

p_2 is the pressure in the higher section

h_2 = 0.250 m is the height in the higher pipe

v_2 is hte speed in the higher section

We can re-write the equation as

v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho} (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-section of the lower pipe, with

r_1 = 3.00 cm =0.03 m is the radius of the lower pipe (half the diameter)

A_2 = \pi r_2^2 is the cross-section of the higher pipe, with

r_2 = 1.50 cm = 0.015 m (radius of the higher pipe)

So we get

r_1^2 v_1 = r_2^2 v_2

And so

v_2 = \frac{r_1^2}{r_2^2}v_1 (2)

Substituting into (1), we find the speed in the lower section:

v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s

B)

Now we can use equation (2) to find the speed in the lower section:

v_2 = \frac{r_1^2}{r_2^2}v_1

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s

C)

The volume flow rate of the water passing through the pipe is given by

V=Av

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume  flow rate is constant, so

r_1=0.03 cm

v_1=0.638 m/s

Therefore, the volume flow rate is

V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0
3 years ago
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