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3 years ago

WHAT are the highest frequency and the lowest frequency pets of the EM spectrum?

Physics

1 answer:

Rus_ich [418]3 years ago

8 0

Highest frequency EM waves: cosmic rays and gama rays

Lowest frequency EM waves:
Radio and Tv waves

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Convert 8 light years to Astronomical Units

marusya05 [52]

Answer:

505929 AU

Explanation:

As you may know, one light-year is equivalent to approximately 63241.1 Astronomical Units. To get your answer, simply multiply 63241.1 * 8 to get ≈505929 AU

5 0

3 years ago

3. Jeremy has a solution with a mass of 228 g. He knows that the mass of the solvent was 224 g. How much is the mass of the solu

mariarad [96]

228 - 224 = 4

there is 4g of solute in the solution.

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2 years ago

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UkoKoshka [18]

Answer:

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3 years ago

A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = $85\frac{1000}{3600}=23.61\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s$

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s$

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2$

Emergency acceleration is 2.99 m/s² (magnitude)

6 0

3 years ago

What frequency fapproach is heard by a passenger on a train moving at a speed of 18.0 m/s relative to the ground in a direction

Sergio039 [100]

Answer:

The frequency is 302.05 Hz.

Explanation:

Given that,

Speed = 18.0 m/s

Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .

We need to calculate the frequency

Using formula of frequency

$f'=f(\dfrac{v+v_{p}}{v-v_{s}})$

Where, f = frequency

v = speed of sound

$v_{p}$ = speed of passenger

$v_{s}$ = speed of source

Put the value into the formula

$f'=262\times(\dfrac{344+18}{344-30})$

$f'=302.05\ Hz$

Hence, The frequency is 302.05 Hz.

7 0

3 years ago