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2 years ago

A copper wire of resistivity 2.6 × 10-8 Ω m, has a cross sectional area of 35 × 10-4 cm2

Physics

1 answer:

KengaRu [80]2 years ago

6 0

Answer:

the length of the wire is 134.62 m.

Explanation:

Given;

resistivity of the copper wire, ρ = 2.6 x 10⁻⁸ Ωm

cross-sectional area of the wire, A = 35 x 10⁻⁴ cm² = ( 35 x 10⁻⁴) x 10⁻⁴ m²

resistance of the wire, R = 10Ω

The length of the wire is calculated as follows;

$R = \frac{\rho L}{A} \\\\L = \frac{RA}{\rho} \\\\L= \frac{10 \times (35\times 10^{-4}) \times 10^{-4}}{2.6 \times 10^{-8}} \\\\L = 134.62 \ m$

Therefore, the length of the wire is 134.62 m.

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The frequency of a wave is 560 Hz. What is it’s period

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The answer would be, "1/560 seconds".

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3 years ago

If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done

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Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

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time in which the power was exerted, t = 1 minute = 60 s.

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Therefore, the work done by the lawnmower is 236.14 J.

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3 years ago

A 10.0kg water balloon is dropped from a height of 12.0m. Calculate the speed of the balloon just before it hits the ground

kolbaska11 [484]

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15.5 m/s.

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Potential energy of the balloon has been converted to kinetic energy.

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5 0

2 years ago

A flutist assembles her flute in a room where the speed of sound is 342 m/sm/s . When she plays the note A, it is in perfect tun

dezoksy [38]

Answer:

A) beats per second she will hear if she now plays the note A as the tuning fork is sounded = 5.13 beats per second

B) length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork = 0.0045m

Explanation:

A) First of all, wavelength = v/f

Where v is speed of wave and f is frequency.

Thus, wavelength of the sound wave of Note A is;

f2 = 440 Hz and v = 342m/s

λ = 342/440 = 0.7773m

Now, since the air inside the note was warmed after a while, the wave will will have a new frequency which we'll call (f1) and and new speed (v'), thus;

f2 = v'/λ = 346/0.7773 = 445.13 Hz

Now let's calculate beat frequency(fbeat).

fbeat = (f1 - f2)

So fbeat = 445.13 - 440 = 5.13Hz or 5.13 beats per second

B) Now, frequency of standing wave models (fm) = n(v/2L)

Where n is a positive integer and L is the open tube length

Making L the subject of the formula, we have; L = nv/2fm

Now from earlier derivation, we see that v = fλ and in this case, v=fλ

Thus, let's replace v with fλ to het;

L = nλ/2

If we take, n=1, L = (1 x 0.7773)/2 = 0.3887m

Now, when the air inside the tube has warmed, it will have a new length to eliminate beats and give same frequency of 440Hz.

So let's call this new length L1;

So L1 = v'/2(f2) = 346/(2x440) = 346/880 = 0.3932m

So the length she needs to extend the "tuning joint" of her flute to be in tune with the tuning fork will be;

ΔL = L1 - L = 0.3932 - 0.3887 = 0.0045m

5 0

3 years ago