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erastova [34]
3 years ago
7

A wave is moving at 18 m/s. If its wavelength is 3 meters, what is its frequency?

Physics
1 answer:
Vlada [557]3 years ago
7 0

here's the solution,

we know that,

=》

wave \: speed = wavelength \times frequency

so,

=》

18 = 3 \times f

=》

f =  \dfrac{18}{3}

=》

f = 6

frequency = 6 hertz

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Lightsail-2 is a spacecraft launched in June 2019 by the Planetary Society driven by radiation pressure on a solar sail of area
lozanna [386]

Answer:

2.655\times 10^{13}\ photons

Explanation:

Area of the solar sail = A = 30 m2

Solar constant = I = 1388 W/m2

Planck's constant = h = 6.626 × 10⁻³⁴ m²kg/s

Speed of light = c = 3×10⁸ m/s

Wavelength of light = \lambda = 570 nm

Pressure from radiation

P=2\frac{I}{c}\\\Rightarrow P=2\frac{1388}{3\times 10^8}\\\Rightarrow P=9.253\times 10^{-6}\ W

Energy of a photon

E=\frac{hc}{\lambda}\\\Rightarrow E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{570\times 10^{-9}}\\\Rightarrow E=3.487\times 10^{-19}\ J

Number of photons

n=\frac{P}{E}\\\Rightarrow n=\frac{9.253\times 10^{-6}}{3.487\times 10^{-19}}\\\Rightarrow n=2.655\times 10^{13}\ photons

Number of photons is 2.655\times 10^{13}\ photons

3 0
3 years ago
A pulley with a radius of 3.0 cm and a rotational inertia of 4.5 x 10–3 kg∙m2 is suspended from the ceiling. A rope passes over
vekshin1

Answer:

22J

Explanation:

Given :

radius 'r'= 3cm

rotational inertia 'I'=4.5 x 10^{-3} kgm²

mass on one side of rope 'm_{1'= 2kg

mass on other side of rope'm_{2' =4kg

velocity'v' of mass m_{2' = 2m/s

Angular velocity of the pulley is given by

‎ω = v /r => 2/ 3x 10^{-2

‎ω = 66.67 rad/s

For the rotating body, we have

KE = \frac{1}{2} I ω²

KE_p = \frac{1}{2} (4.5 *10^{-3} )(66.67^{2} )

KE_p = 10J

Next is to calculate kinetic energy of the blocks :

KE_{b} = \frac{1}{2} (m_1 + m_2).v^2\\KE_b= \frac{1}{2} (2+4).2^2

KE_b=12J

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KE= 22J

6 0
3 years ago
What does it mean for waves to be polarized?
Yuki888 [10]

Answer:

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Explanation:

Hope this helps!!!!!

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4 0
3 years ago
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Your ear is capable of differentiating sounds that arrive at each ear just 0.34 ms apart, which is useful in determining where l
goblinko [34]

Answer:

Δt = 5.29 x 10⁻⁴ s = 0.529 ms

Explanation:

The simple formula of the distance covered in uniform motion can be used to find the interval between when the sound arrives at the right ear and the sound arrives at the left ear.

\Delta s = v\Delta t\\\\\Delta t = \frac{\Delta s}{v}

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Δt = required time interval = ?

Δs = distance between ears = 18 cm = 0.18 m

v = speed of sound = 340 m/s

Therefore,

\Delta t = \frac{0.18\ m}{340\ m/s}

<u>Δt = 5.29 x 10⁻⁴ s = 0.529 ms</u>

4 0
3 years ago
Should you be worried if a n eighteen wheeler truck is riding close behind your car? Explain why or why not. Please help me D:
xz_007 [3.2K]
For some reasons, no. If the driver looks focused and has experience, then it would be okay. Again, it could be dangerous if you bump into the truck, it would cause damage to you and your passengers. 

Mostly, I would agree with 'No'. :)
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3 years ago
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