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Gnoma [55]
4 years ago
7

Wire A and wire B are made from the same material and are under the same tension, but waves travel along string B at one-third t

he wave speed of string A. The radius of wire A is 0.100 mm. What is the radius of wire B
Physics
1 answer:
Ahat [919]4 years ago
4 0

Answer:

0.3 mm

Explanation:

\rho = Density of material

\mu = Linear density of the material = \rho A

Area = Area = \pi r^2

r = Radius

The velocity of a wave is given by

v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{T}{\rho A}}\\\Rightarrow v=\sqrt{\dfrac{T}{\rho \pi r^2}}

It can be seen that the velocity is inversly proportional to the radius

v\propto \sqrt{\dfrac{1}{r^2}}\\\Rightarrow v\propto \dfrac{1}{r}

So,

\dfrac{v_a}{v_b}=\dfrac{r_b}{r_a}

From the question

v_b=\dfrac{1}{3}v_a

\\\Rightarrow \dfrac{v_a}{\dfrac{1}{3}v_a}=\dfrac{r_b}{0.1}\\\Rightarrow 3=\dfrac{r_b}{0.1}\\\Rightarrow r_b=3\times 0.1\\\Rightarrow r_b=0.3\ mm

The radius of wire B is 0.3 mm

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Answer:

a) i = -9.63 cm ,    h ’= .0.24075 cm   erect

b)  i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

               1 / f = (n-1) (1 / R₁ - 1 / R₂)

                1 / f = (1.70 -1)) 1 / ∞ - 1/13)

                1 / f = 0.0538

                 f = - 18.57 cm

Now we can use the constructor equation

             1 / f = 1 / o + 1 / i

             1 / i = 1 / f - 1 / o

              1 / i = -1 / 18.57 -1/20

               1 / i = -0.1038 cm

               I = -9.63 cm

For the height of the

image let's use magnification

                 m = h '/ h = - i / o

                  h ’= -h i / o

                  h ’= - 0.5 (-9.63) / 20

                  h ’= .0.24075 cm

b) we invert the lens

The focal length is

             1 / f = (1.70 -1) (1/13 - 1 / int)

              1 / f = 0.0538

             f = 18.57 cm

             1 / i = 1 / f -1 / o

             1 / I = 1 / 18.57 - 1/20

             1 / I = 3.85 10-3

             i = 259.74 cm

     

            h ’= - 0.5 259.74 / 20

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Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
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Answer:

a). \frac{\dot{W}}{m}= 311 kJ/kg

b). \frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0

\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})

\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})

\frac{\dot{W}}{m}= -30+1.1(980-670)

\frac{\dot{W}}{m}= 311 kJ/kg

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\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0

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A note of frequency 200Hz has a velocity of 400m/s. what is the wavelength of the note​
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Answer:

\huge\boxed{\sf \lambda = 2 m}

Explanation:

<h3><u>Given data:</u></h3>

Frequency = f = 200 Hz

Velocity = v = 400 m/s

<h3><u>Required:</u></h3>

Wavelength = λ = ?

<h3><u>Formula:</u></h3>

v = fλ

<h3><u>Solution:</u></h3>

Put the givens in the formula

400 = (200)λ

Divide 200 to both sides

400/200 = λ

2 m = λ

λ = 2 m

\rule[225]{225}{2}

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2 years ago
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