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3 years ago

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A stationary source S generates circular outgoing waves on a lake. The wave speed is 5.0 m/s and the crest-to-crest distance is

2.0 m. A person in a motorboat heads directly toward S at 3.0 m/s. To this person, the frequency of these waves is:

1 answer:

Aleonysh [2.5K]3 years ago

4 0

Answer:4 Hz

Explanation:

Speed of wave $v=5 m/s$

crest to crest distance $\lambda =2 m$

velocity of observer $v_0=3 m/s$

actual frequency $f=\frac{velocity}{\lambda }$

$f=\frac{5}{2}=2.5 Hz$

Apparent frequency $f'=f(\frac{v+v_0}{v})$

$f'=2.5\times \frac{5+3}{5}$

$f'=4 Hz$

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Answer:

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Explanation

Let A the total area of the galaxy, is modeled as a disc:

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And let a be the area that astronomers are able to see:

a = πr^2 = π(5 kpc)^2

The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:

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Read 2 more answers

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

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Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

So

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P_f = k * \frac{q_1 * q_2 }{R_f } - 0$

So

$\frac{1}{2} * m * \frac{1}{8} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R_f }$

=> $\frac{1}{2} * m *v_o^2 [-\frac{15}{16} ] = k * \frac{q_1 * q_2 }{R_f }$

=> $- \frac{15}{32} * m *v_o^2 = k * \frac{q_1 * q_2 }{R_f } ---(2)$

Divide equation 2 by equation 1

$\frac{- \frac{15}{32} * m *v_o^2 }{- \frac{3}{8} * m *v_0^2 } } = \frac{k * \frac{q_1 * q_2 }{R_f } }{k * \frac{q_1 * q_2 }{R } }}$

=> $-\frac{15}{32 } * -\frac{8}{3} = \frac{R}{R_f}$

=> $\frac{5}{4} = \frac{R}{R_f}$

=> $R_f = \frac{4}{5} R$

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3 years ago

A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they pr

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A) Order of the first laser: 3, order of the second laser: 2

B) The overlap occurs at an angle of $34.9^{\circ}$

Explanation:

A)

The formula that gives the position of the maxima (bright fringes) for a diffraction grating is

$d sin \theta = m \lambda$

where

d is spacing between the lines in the grating

$\theta$ is the angle of the maximum

m is the order of diffraction

$\lambda$ is the wavelength of the light

For laser 1,

$d sin \theta = m_1 \lambda_1$

For laser 2,

$d sin \theta = m_2 \lambda_2$

where

$\lambda_1 = 420 nm\\\lambda_2 = 630 nm$

Since the position of the maxima in the two cases overlaps, then the term $d sin \theta$ on the left is the same for the two cases, therefore we can write:

$m_1 \lambda_1 = m_2 \lambda_2\\\frac{m_1}{m_2}=\frac{\lambda_2}{\lambda_1}=\frac{630}{420}=\frac{3}{2}$

Therefore:

$m_1 = 3$

$m_2 = 2$

B)

In order to find the angle at which the overlap occurs, we use the 1st laser situation:

$d sin \theta = m_1 \lambda_1$

where:

N = 450 lines/mm = 450,000 lines/m is the number of lines per unit length, so the spacing between the lines is

$d=\frac{1}{N}=\frac{1}{450,000}=2.2\cdot 10^{-6} m$

$m_1 = 3$ is the order of the maximum

$\lambda_1 = 420 nm = 420\cdot 10^{-9} m$ is the wavelength of the laser light

Solving for $\theta$ , we find the angle of the maximum:

$sin \theta = \frac{m_1 \lambda_1}{d}=\frac{(3)(420\cdot 10^{-9})}{2.2\cdot 10^{-6}}=0.572$

So the angle is

$\theta=sin^{-1}(0.572)=34.9^{\circ}$

Learn more about diffraction:

brainly.com/question/3183125

#LearnwithBrainly

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3 years ago