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Helen [10]
3 years ago
8

What is the wavelength (in air) of a 13.0 khz vlf radio wave?

Physics
1 answer:
bonufazy [111]3 years ago
5 0
Since,
Speed = Frequency * WaveLength

=> WaveLength = Speed / Frequency --- (A)

Frequency = 13.0 kHz.
As the radio waves are electromagnetic waves, their speed is equals to the speed of light. Therefore,
Speed = C = 3 *  10^{8}  m/ s^{2}

Plug in the values in equation(A):

A => WaveLength = \frac{3 * 10^{8}}{13*10^{3}}

Ans: Wavelength = 23.077 kilometers.
-i
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he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug
S_A_V [24]

Answer:

207.4 N

Explanation:

The torque \tau  on a body is

\tau = r* F  where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

\tau = rF\sin \theta

Where \theta is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation \tau = rF\sin \theta

\tau = LF\sin \theta

Where L is the length of the wrench.

Making F the subject

F = \frac{\tau }{{L\sin \theta }}

Force required to pull the wrench is given as,

F = \frac{\tau }{{L\sin \theta }}

Substitute 47{\rm{ N}} \cdot {\rm{m}}  for \tau, 25 cm for L, and 115o for \theta  

\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}  

6 0
2 years ago
Sound that reaches the ears after bouncing off a wall or a floor is called _____.
andrew11 [14]

Answer:

Indirect sound

8 0
1 year ago
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In any energy transformation, there is always some energy that gets wasted as non-useful heat.
Nady [450]
It is a completely false statement that in <span>any energy transformation, there is always some energy that gets wasted as non-useful heat. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
8 0
3 years ago
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The acceleration due to gravity is g at the earth's surface. an earth satellite is orbiting at a distance from the earth's surfa
disa [49]

The acceleration due to gravity is g/4

The acceleration above the earth surface is given by the relation

g^'=gr^2/〖(h+r)〗^2

Since the satellite orbits the earth in a orbit of radius equal to earth radius, therefore

g^'=(gr^2)/〖(r+r)〗^2 =g/4

Thus the acceleration due to gravity on the satellite is g/4.

6 0
3 years ago
In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. if the passengers are to ex
sleet_krkn [62]
The answer is "156.6 m/s".

This is how we calculate this;

-N + mg = ma = mv²/r

For "weightlessness" N = 0, so

0 = mg - mv²/r 

g - v²/r = 0 

v =√( gr)
g = 9.8 and r = 2.5km = 2500 m

v = √(9.8 x 2500)

= 156.6 m/s
8 0
3 years ago
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