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3 years ago

7

How would you classify an EM wave with a frequency of 10^7 Hz?

1 answer:

tiny-mole [99]3 years ago

5 0

Answer:

I think it is television and radio wave

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An experiment yielded the above temperature and time information. what is the freezing point of the material in this experiment

Dmitry [639]

The answer is 0 degrees Celsius (0°C). It will be where the line flat lines the first time. The second time would be the boiling point. An experiment yielded the above temperature and time information. The freezing point of the material in this experiment if the material is a solid at time zero is 0 degrees Celsius (0°C) .

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3 years ago

An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

SCORPION-xisa [38]

Answer:

$a_y=0.92m/s^2$

Explanation:

To solve this problem we use the formula for accelerated motion:

$y=y_0+v_{y0}t+\frac{a_yt^2}{2}$

We will take the initial position as our reference ( $y_0=0m$ ) and the downward direction as positive. Since the rock departs from rest we have:

$y=\frac{a_yt^2}{2}$

Which means our acceleration would be:

$a_y=\frac{2y}{t^2}$

Using our values:

$a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2$

5 0

3 years ago

If your Stronger than Hisoka, and you know it, Clap your hands! Oh, wait.. That’s right! You don’t have any hands. Lol

ddd [48]

Answer:

Haha pretty funny

Explanation:

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2 years ago

Read 2 more answers

The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of th

Lady_Fox [76]

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

$=10^{-5.9}$

intensity of sound of 5 persons

$I=5\times 10^{-5.9}$

$dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}$

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

4 0

3 years ago

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago