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3 years ago

“Is it correct to say that a radio wave can be considered a low-frequency light wave?

1 answer:

4 0

It's weird but technically correct to say that a radio wave can be considered a low-frequency light wave. Radio and light are both electromagnetic waves. The only difference is that radio waves have much much much longer wavelengths, and much much much lower frequencies, than light waves have. But they're both the same physical phenomenon.

However, a radio wave CAN'T also be considered to be a sound wave. These two things are as different as two waves can be.

-- Radio is an electromagnetic wave. Sound is a mechanical wave.

-- Radio waves travel more than 800 thousand times faster than sound waves do.

-- Radio waves are transverse waves. Sound waves are longitudinal waves.

-- Radio waves can travel through empty space. Sound waves need material stuff to travel through.

-- Radio waves can be detected by radio, TV, and microwave receivers. Sound waves can't.

-- Sound waves can be detected by our ears. Radio waves can't.

-- Sound waves can be generated by talking, or by hitting a frying pan with a spoon. Radio waves can't.

-- Radio waves can be generated by an alternating current flowing through an isolated wire. Sound waves can't.

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A Shaolin monk of mass 60 kg is able to do a ‘finger stand’: he supports his whole weight on his two index fingers, giving him a

LekaFEV [45]

Answer:

1500000 Pa

Explanation:

The formula for pressure is force per unit area.

P=F/A where F is force and A is area

Given that ;

F= mass * acceleration due to gravity

F= 60 * 9.81 = 588.6 = 589 N

A= area = 4cm² = 0.0004 m²

P= F/A = 589 / 0.0004

P= 1471500

P=1500000 Pa

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3 years ago

Ag,Au and Cu are called coinage metals why plzzzz hurry its urgent plzzz

Ronch [10]

Answer:

This is because these metals are used for minting (making) coins.

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3 years ago

The magnetic field at the centre of a toroid is 2.2-mT. If the toroid carries a current of 9.6 A and has 6.000 turns, what is th

ziro4ka [17]

Answer:

Radius, r = 0.00523 meters

Explanation:

It is given that,

Magnetic field, $B=2\ mT=2.2\times 10^{-3}\ T$

Current in the toroid, I = 9.6 A

Number of turns, N = 6

We need to find the radius of the toroid. The magnetic field at the center of the toroid is given by :

$B=\dfrac{\mu_oNI}{2\pi r}$

$r=\dfrac{\mu_oNI}{2\pi B}$

$r=\dfrac{4\pi \times 10^{-7}\times 6\times 9.6}{2.2\pi \times 2\times 10^{-3}}$

r = 0.00523 m

$r=5.23\times 10^{-3}\ m$

So, the radius of the toroid is 0.00523 meters. Hence, this is the required solution.

7 0

3 years ago

Read 2 more answers

Pls help will mark Brainliest

BartSMP [9]

Answer: 100 units

Explanation:

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3 years ago

A Ferris wheel on a California pier is 27 m high and rotates once every 32 seconds in the counterclockwise direction. When the w

Leto [7]

A) 140 degrees

First of all, we need to find the angular velocity of the Ferris wheel. We know that its period is

T = 32 s

So the angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{32 s}=0.20 rad/s$

Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

$\theta= \omega t$

and substituting t = 75 seconds, we find

$\theta= (0.20 rad/s)(75 s)=15 rad$

In degrees, it is

$15 rad: x = 2\pi rad : 360^{\circ}\\
x=\frac{(15 rad)(360^{\circ})}{2\pi}=860^{\circ} = 140^{\circ}$

So, the new position is 140 degrees from the initial position at the top.

B) 2.7 m/s

The tangential speed, v, of a point at the egde of the wheel is given by

$v=\omega r$

where we have

$\omega=0.20 rad/s$

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

$v=(0.20 rad/s)(13.5 m)=2.7 m/s$

6 0

3 years ago