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1 year ago

A transverse wave on a string is described by the wave function

Physics

1 answer:

svp [43]1 year ago

5 0

The period of the transverse wave from what we have here is 0.5

<h3>How to find the period of the transverse wave</h3>

The period of a wave can be defined as the time that it would take for the wave to complete one complete vibrational cycle.

The formula with which to get the period is

w = 4π

where w = 4 x 22/7

2π/T = 4π

6.2857/T = 12.57

From here we would have to cross multiply

6.2857 = 12.57T

divide through by 12.57

6.2857/12.57 = T

0.500 = T

Hence we can conclude that the value of T that can determine the period based on the question is 0.500.

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A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

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3 years ago

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vladimir1956 [14]

<span>What is the main fuel consumed in the core of a red giant?
The </span><span>main fuel consumed in the core of a red giant is He or helium. The answer is letter D.</span>

4 0

3 years ago

A locomotive engine pulls a train with a constant force comment on whether its acceleration will increase or decrease if more co

julia-pushkina [17]

Explanation:

the acceleration will be unchanged according to newton second law of motion

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3 years ago

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6 0

3 years ago

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl

USPshnik [31]

Given Information:

Wavelength = λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

3 0

3 years ago