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Mnenie [13.5K]
3 years ago
11

A system proposed by the U.S. Navy for underwater submarine communication, called ELF (for extremely low frequency), operates wi

th a frequency of 30Hz . What is the wavelength of this radiation in meters
Physics
2 answers:
kow [346]3 years ago
7 0

Answer

Wavelength= 30*20^8/30=10^7m

Explanation:

Velocity = frequency *wavelength

We're frequency=30HZ

Velocity of light= 3*10^8m/s

Wavelength= 30*20^8/30=10^7m

Explanation:

Nesterboy [21]3 years ago
5 0

Answer:

the wavelength of this radiation is 10 × 10⁶m = 10⁷m

Explanation:

Frequency, v = 30Hz = 30s⁻¹

wavelength of radiation = λ

Wavelength = speed of light / frequency

c = speed of light

c = 3 × 10⁸m/s

v = frequency

Wavelength =  3 × 10⁸ / 30

Wavelength = 10000000

Wavelength = 10 × 10⁶m

Wavelength = 10⁷m

Hence,  the wavelength of this radiation is 10 × 10⁶m = 10⁷m

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How many significant figures does the following number have: 0.002040?
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Answer: 4

Explanation: because 0s aren’t significant and after the decimal point, there was to be a value greater than 0 than the rest are sig figs.

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2 years ago
What are the differences between refraction and diffraction?
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Refraction is the change in direction of a wave.
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5 0
3 years ago
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
Solved, can someone check it over! ​
kakasveta [241]

Answer:

its good no need to change anything :))

4 0
2 years ago
Which claim would most likely be considered valid?
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Please state the options and I will answer to the best of my abilities XD
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3 years ago
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