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sweet-ann [11.9K]

3 years ago

9

A dolphin emits a sound wave that hits a target 120 m away. The wave reflects back from the target to the dolphin. If the Bulk m

odulus of seawater is 2.3 x 10° N/m2 and the density of seawater is 1022 kg/m3. How long does it take the sound wave for the round trip?

1 answer:

Veronika [31]3 years ago

3 0

Answer:

The time taken for the sound wave to make the round trip is 0.16 s.

Explanation:

Given;

distance traveled by the sound wave, d = 120 m

bulk modulus of sea water, B = 2.3 x 10⁹ N/m²

density of sea water, ρ = 1022 kg

The speed of the wave is given by;

$v = \sqrt{\frac{B}{\rho} } \\\\v = \sqrt{\frac{2.3*10^9}{1022} }\\\\v = 1500.16 \ m/s$

Speed is given by;

$Speed = \frac{Distance}{Time}$

total distance of the round trip = 2 x 120m = 240 m

Time taken for the sound wave to make the round trip is given by;

$Time = \frac{Distance}{Speed} \\\\Time = \frac{240}{1500.16} \\\\Time = 0.16 \ second$

Therefore, the time taken for the sound wave to make the round trip is 0.16 s.

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A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

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