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2 years ago

6

The variation in the pressure of helium gas, measured from its equilibrium value, is given by ΔP = 2.9 × 10−5 cos (6.20x − 3 000

t), where x and t have units m and s, and ΔP is measured in N/m2. Determine the wavelength (in m) of the wave.

1 answer:

nadya68 [22]2 years ago

6 0

Answer:

The wavelength of this wave is 1.01 meters.

Explanation:

The variation in the pressure of helium gas, measured from its equilibrium value, is given by :

$\Delta P=2.9\times 10^{-5}\ cos(6.2x-3000t)$ ..............(1)

The general equation is given by :

$\Delat P=P_o\ cos(kx-\omega t)$ ...........(2)

On comparing equation (1) and (2) :

$k=6.2$

Since, $k=\dfrac{2\pi}{\lambda}$

$\dfrac{2\pi}{\lambda}=6.2$

$\lambda=1.01\ m$

So, the wavelength of this wave is 1.01 meters. Hence, this is the required solution.

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Explanation:

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2 years ago

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

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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:

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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:

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

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

2 years ago