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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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Name one metal and one non-metal from the periodic table? <br><br> Y’all I need help<br> Thank you!
Stella [2.4K]

Answer:

Zinc and Neon!

Explanation:

zinc is an element that is a transition metal and neon is a noble gas

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What phenomena provides <br> evidence of electric fields in the atmosphere
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The phenomena<span> of </span>atmospheric<span> electricity are of three kinds. ..... In the Earth-</span>ionosphere cavity, the electric field<span> and conduction current in the lower </span>atmosphere<span> </span>
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Please Help!
irga5000 [103]

The maximum force that the tires can exert on the road before slipping is 16200 N.

From the information in the question;

The coefficient of static friction =  0.9

The mass of the car = 1800 kg

Using the formula;

μ = F/R

μ  = coefficient of static friction

F = force on the tires

R = the reaction force

But recall that the reaction is equal in magnitude to the weight of the car.

W=R

Hence; R = 1800 kg × 10 ms-2 = 18000 N

Making F the subject of the formula;

F = μR

Substituting values;

F =  18000 N × 0.9

F = 16200 N

Hence, the maximum force that the tires can exert on the road before slipping is 16200 N.

Learn more: brainly.com/question/18754989

6 0
2 years ago
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Answer:

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6 0
2 years ago
Bill and Ted are standing on a bridge 40 ft above a river. Bill drops a stone, while Ted decides to throw a stone downward at 10
USPshnik [31]

Answer:

D.

Explanation:

In order to know how long after Bill released his rock should Ted throw his if they want the stones to hit the water simultanously, we need to calculate the time needed to hit the water to both rocks independent each other, and just take the difference.

For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

y = \frac{1}{2} * g * t^{2}

where y = height = 40 ft.

As all the parameters are given in SI units, it is  advisable to convert this value to m, as follows:

y = 40 ft*\frac{0.3048m}{1 ft} = 12.2 m

Now, we can solve for t, as follows:

t = \sqrt{\frac{2*y}{g}} =  \sqrt{\frac{2*12.2m}{9.8m/s2}} = 1.58 s

For the rock thrown down at 10 m/s, the kinematic equation we just have used becomes:

y = v0*t +\frac{1}{2} * g * t^{2}

This leaves us a quadratic equation on t, as follows:

t = \frac{-10m/s}{9.8m/s2} +/- \sqrt{10m/s^{2} -4*4.9m/s2*(-12.2m)} = -1.02 s +/- 1.88s

Taking the positive root, we have:

t = -1.02 s + 1.88 s = 0.86 s

So, in order to get that both rocks hit the water at the same time, Ted will need to wait the difference between both times:

Δt = 0.86 s - 1.58s = -0.72 s

5 0
3 years ago
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