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2 years ago

12

Two sinusoidal waves are identical except for their phase. When these two waves travel along the same string, for which phase di

fference will the amplitude of the resultant wave be a maximum?

1 answer:

Kamila [148]2 years ago

6 0

Answer:

zero or 2π is maximum

Explanation:

Sine waves can be written

x₁ = A sin (kx -wt + φ₁)

x₂ = A sin (kx- wt + φ₂)

When the wave travels in the same direction

Xt = x₁ + x₂

Xt = A [sin (kx-wt + φ₁) + sin (kx-wt + φ₂)]

We are going to develop trigonometric functions, let's call

a = kx + wt

Xt = A [sin (a + φ₁) + sin (a + φ₂)

We develop breasts of double angles

sin (a + φ₁) = sin a cos φ₁ + sin φ₁ cos a

sin (a + φ₂) = sin a cos φ₂ + sin φ₂ cos a

Let's make the sum

sin (a + φ₁) + sin (a + φ₂) = sin a (cos φ₁ + cos φ₂) + cos a (sin φ₁ + sinφ₂)

to have a maximum of the sine function, the cosine of fi must be maximum

cos φ₁ + cos φ₂ = 1 +1 = 2

the possible values of each phase are

φ1 = 0, π, 2π

φ2 = 0, π, 2π,

so that the phase difference of being zero or 2π is maximum

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sesenic [268]

Work done by a given force is given by

$W = F.d$

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = $W_1 = Fdcos\theta$

$W_1 = 12*5cos15$

$W_1 = 57.96 J$

Now similarly work done by frictional force

$W_2 = Fdcos\theta$

$W_2 = 4*5cos180$

$W_2= -20 J$

Now total work done on sled

$W_{net}= W_1 + W2$

$W_{net} = 57.96 - 20 = 37.96 J$

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2 years ago

In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053nm around a stationary proton. par

Alenkasestr [34]

<span>Here the force that is applied between the electron and proton is centripetal, so equate the two forces to determine the velocity.
We know charge of the electron which for both Q1 and Q2, e = 1.60 x 10^-19 C
The Coulombs Constant k = 9.0 x 10^9
Radius r = 0.053 x 10^-9m = 5.3 x 10^-11 m
Mass of the Electron = 9.11 x 10^-31
F = k x Q1 x Q2 / r^2 = m x v^2 / r(centripetal force)
ke^2 / r^2 = m x v^2 / r => v^2 = ke^2 / m x r
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v^2 = 4.77 x 10^12 = 2.18 x 10^6 m/s
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one orbit = circumference = 2 x pi x r; distance s = v x t.
v x t = 2 x pi x r => t = (2 x 3.14 x 5.3 x 10^-11) / (2.18 x 10^6)
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2 years ago

If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

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2 years ago

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Korolek [52]

Answer:

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Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒ $I_1\omega_1 = I_2\omega_2$

$\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}$

also,

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since, $r_2^2$

$I_2^2$

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$\omega_1^2$

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2 years ago

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Nadusha1986 [10]

Loudness is one of answers

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2 years ago