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2 years ago

Calculate the number of electrons passing a point in the wire in 1 min when the current is 1 A

Physics

1 answer:

Vera_Pavlovna [14]2 years ago

7 0

Explanation:

When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017

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How is it possible to get three precise but inaccurate measurements of the same volume of water

Lubov Fominskaja [6]

The inaccurate measurements must be similar to the other two measurements (ex; 590, 589, 599), but different from the actual volume of water. (Ex; the actual volume is let say.. 100, but you measured 50, 49, 40)

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2 years ago

Ayuda por favor (archivo adjunto) con un ejercicio de expresión sobre periodo de oscilación de esta figura:

vodka [1.7K]

Answer:

nolo se

Explanation:

no lo se

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2 years ago

How is the acceleration of an object in uniform circular motion constant?

nevsk [136]

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3 years ago

Read 2 more answers

A child has an ear canal that is 1.3 cm long. Assume the speed of sound is v = 344 m/s.

kap26 [50]

Answer:

The frequencies are $(f, f_1) = (6615.4 \ Hz , 19846.2\ Hz)$

Explanation:

From the question we are told that

The length of the ear canal is $l = 1.3 \ cm =\frac{1.3}{100} = 0.013 \ m$

The speed of sound is assumed to be $v_s = 344 \ m/s$

Now taking look at a typical ear canal we see that we assume it is a closed pipe

Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as

$f = \frac{v_s}{4 * l }$

substituting values

$f = \frac{344}{4 * 0.013 }$

$f = 6615.4 \ Hz$

Also the the second harmonic for the pipe (ear canal) is mathematically represented as

$f_1 = \frac{3v_s}{4 * l}$

substituting values

$f_1 = \frac{3 * 344}{4 * 0.013}$

$f_1 = 19846.2 \ Hz$

Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies

6 0

3 years ago

An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Out

Sphinxa [80]

Answer:

The first minimum would be observed at 41.57°

Explanation:

v = 340m/s = speed of sound

f = 610Hz

d = 0.840m

λ = ?

Mλ = wsinθ

m = mth order minima

λ = wavelength incident on the single slit

θ = angular position of the mth minima

But, λ = v / f

λ = 340 / 610 = 0.557m

θ = sin⁻(mλ/d)

θ = sin⁻ [(1 * 0.557) / 0.840]

θ = sin⁻ 0.6635

θ = 41.57°

The first minimum would be observed at 41.57°

4 0

3 years ago