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sergeinik [125]
3 years ago
11

Edit question A 37-cm-long wire of linear density 18 g/m vibrating at its second mode, excites the third vibrational mode of a t

ube of length 192 cm that is open near the wire and closed at the other end. Please find the tension in the wire assuming that the speed of sound in air is 343 m/s.
Physics
1 answer:
Pachacha [2.7K]3 years ago
7 0

Answer:

176.9N

Explanation:

The following data were given

wire length,L=37cm=0.37m

linear density=18g/m

tube length,=192cm=1.92m,

speed of sound,v=343m/s

Since it is an open-closed tube, the second harmonic frequency is expressed as

f_{3}=3(\frac{v}{4l} )\\f_{3}=3(\frac{343}{4*1.92})\\f_{3}=133.98Hz

The relationship between the tension,  linear density and second harmonic frequency is expressed as

f_{3}=\frac{1}{2l_{w}}\sqrt{\frac{T}{\alpha } } \\T=(f_{3}*2l_{w})^{2}\alpha \\T=(133.984*2*0.37)^{2}*18*10^{-3}\\T=176.9N

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Answer:

1807.56 kJ

Explanation:

Parameters given:

Current, I = 8.9A

Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs

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Electrical energy is given as:

E = I*V*t

Where I = Current

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E = 8.9 * 12 * 16920

E = 1807056 J = 1807.056 kJ

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An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
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Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

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Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

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Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

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Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

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You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta
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Answer:

1)  3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes:    \frac{Distance}{Speed}= \frac{22}{6} =   3.66

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes:    Speed * Time = 34 * 3.66 = 124.44

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : s = u*t + \frac{1}{2}(a * t^2)

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1  = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

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