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4 years ago

What is the frequency of an ocean wave that is traveling at a speed of 45 m/s if it has a wavelength of 3 meters.

Physics

1 answer:

vaieri [72.5K]4 years ago

5 0

Answer:

Frequency, f = 15 Hz

Explanation:

We have,

Speed of an ocean wave is 45 m/s

Wavelength of a wave is 3 m

It is required to find the frequency of an ocean wave.

Speed of a wave, $v=f\lambda$ , f = frequency of ocean wave

$f=\dfrac{v}{\lambda}\\\\f=\dfrac{45}{3}\\\\f=15\ Hz$

So, the frequency of an ocean wave is 15 Hz.

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2 years ago

The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.

Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

The angular velocity of the telescope is approximately 0.199 rad/s

Second question:

The telescope should accelerates away by approximately F = 0.0005·m

Third question:

The pulling force between the Earth and the satellite

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

The angular velocity of the telescope, ω ≈ 0.199 rad/s

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$ , therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ 0.0005·m (away from the Sun)

Third part:

Other forces include;

The force of attraction between the Earth and the telescope which can contribute to the the telescope having a stable orbit at the given speed.

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3 years ago

Science is hard sometimes help please!

dolphi86 [110]

Yeah science is hard. Hopefully you do good in it :)

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3 years ago

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Explanation:

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It is given by the expression

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since kinetic energy is directly proportional to the square of the velocity, the kinetic energy of an object increases with its velocity.

Hence an object that moves at the fastest speed will have the most kinetic energy.

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3 years ago

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allsm [11]

Energy is unable to change forms. please select the best answer from the choices provided

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