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katrin2010 [14]

3 years ago

10

A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s.

(Enter your answers to at least four decimal places.) (a) How far away (in m) is the explosion if air temperature is 22.0°C and if you neglect the time taken for light to reach the physicist? In meters. (b) Calculate the distance to the explosion (in m) taking the speed of light into account. Note that this distance is negligibly greater. In meters.

1 answer:

True [87]3 years ago

4 0

Answer:

a) 137.6 m

b) 137.632 m

Explanation:

T = Temperature of air = 22.0 °C

v = speed of sound at temperature "T"

speed of sound at temperature "T" is given as

$v = 331\sqrt{1 + \frac{T}{273}}$

$v = 331\sqrt{1 + \frac{22}{273}}$

$v = 344.08$ m/s

a)

t = time taken for the sound to reach the physicist = 0.400 s

d = distance of the location of explosion

Distance of the location of explosion is given as

d = v t

d = (344.08) (0.400)

d = 137.6 m

b)

d = distance of the location of explosion

v = speed of sound = 344.08 m/s

c = speed of light = 3 x 10⁸ m/s

$t_{s}$ = time taken by sound to reach the physicist = $\frac{d}{v}$ = $\frac{d}{344.08}$

$t_{L}$ = time taken by light to reach the physicist = $\frac{d}{c}$ = $\frac{d}{3\times 10^{8}}$

t = time taken for the sound to reach the physicist = 0.400 s

time taken for the sound to reach the physicist is given as

t = $t_{s}$ - $t_{L}$

0.400 = $\frac{d}{344.08}$ - $\frac{d}{3\times 10^{8}}$

d = 137.632 m

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Answer:

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The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

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$p = \frac{RA}{L}$

$A = \frac{pL}{R}$

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where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A and λ is use for (m/L) in the eqn above

$\lambda = d\frac{p}{\frac{R}{L} } ------(3)$

Substitute 0.200Ω.km⁻¹ for (R/L)

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$J = \frac{iR}{pL}$

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

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$J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2$

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

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Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

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