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Akimi4 [234]
3 years ago
12

In a thunderstorm at 20.0°C, Karen sees a bolt of lightning and hears the thunderclap 3.00 s later. How far from Karen did the l

ightning strike? Show your work.
Please be sure to show all work. Thank you so much :D
Physics
1 answer:
Minchanka [31]3 years ago
7 0
-- The speed of light in air is very close to 3 x 10⁸ m/s.
Whatever the actual number is, it's equivalent to roughly
7 times around the Earth in 1 second.  So for this kind of
problem, you can assume that we see things at the same time
that they happen; don't bother worrying about how long it takes
for the light to reach you.

-- For sound, it's a different story.  Sound in air only travels at
about 340 m/s.  It takes sound almost 5 seconds to go 1 mile.

-- Now, the lightning and thunder happen at the same time.
The light travels to you at the speed of light, so you see the
lightning pretty much when it happens.  But the sound of the
thunder comes poking along at 340 m/s, and arrives AFTER
the sight of the lightning.

The length of time between the sight and the sound is about
99.9999% the result of the time it takes the sound to reach you.

If the thunder arrived at you 3 seconds after the light did, then
the sound traveled
        
                     (340 m/s) x (3 s) =  1,020 meters .
                                           
(about 0.63 of a mile)

(If you're worried about ignoring the time it takes
for the light to reach you ...

  It takes light  0.0000034 second to cover the same 1,020 meters,

so including it in the calculation would not change the answer.)

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A driver of a car enters a new 110 km/h speed zone on the highway. The driver begins to accelerate immediately and reaches 110 k
levacccp [35]

Answer:

30Km/h

Explanation:

acceleration is the change of speed in a given time so when we substract the accelerations we can know how much the car goes per an hour

3 0
3 years ago
Name any two liquid which are found in our human eyes​
Romashka [77]

Aqueous humor and vitreous humor are the liquids present in the human eye.

<em>Hope </em><em>it</em><em> helped</em><em> you</em><em>.</em><em>.</em><em>.</em><em> </em><em>pls </em><em>mark</em><em> brainliest</em>

8 0
3 years ago
A train whistle is heard at 300 Hz as the train approaches town. The train cuts its speed in half as it nears the station, and t
spin [16.1K]

Answer:

The speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.

Explanation:

We can calculate the speed of the train using the Doppler equation:

f = f_{0}\frac{v + v_{o}}{v - v_{s}}        

Where:

f₀: is the emitted frequency

f: is the frequency heard by the observer  

v: is the speed of the sound = 343 m/s

v_{o}: is the speed of the observer = 0 (it is heard in the town)

v_{s}: is the speed of the source =?

The frequency of the train before slowing down is given by:

f_{b} = f_{0}\frac{v}{v - v_{s_{b}}}  (1)                  

Now, the frequency of the train after slowing down is:

f_{a} = f_{0}\frac{v}{v - v_{s_{a}}}   (2)  

Dividing equation (1) by (2) we have:

\frac{f_{b}}{f_{a}} = \frac{f_{0}\frac{v}{v - v_{s_{b}}}}{f_{0}\frac{v}{v - v_{s_{a}}}}

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - v_{s_{b}}}   (3)  

Also, we know that the speed of the train when it is slowing down is half the initial speed so:

v_{s_{b}} = 2v_{s_{a}}     (4)

Now, by entering equation (4) into (3) we have:

\frac{f_{b}}{f_{a}} = \frac{v - v_{s_{a}}}{v - 2v_{s_{a}}}  

\frac{300 Hz}{290 Hz} = \frac{343 m/s - v_{s_{a}}}{343 m/s - 2v_{s_{a}}}

By solving the above equation for v_{s_{a}} we can find the speed of the train after slowing down:

v_{s_{a}} = 11.06 m/s

Finally, the speed of the train before slowing down is:

v_{s_{b}} = 11.06 m/s*2 = 22.12 m/s

Therefore, the speed of the train before and after slowing down is 22.12 m/s and 11.06 m/s, respectively.                        

I hope it helps you!                                                        

7 0
3 years ago
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op
Juliette [100K]

Answer:

The fundamental frequency of can is 2.7 kHz.                          

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

7 0
3 years ago
which of the following cannot be determined by looking at the phase diagram? A. melting point, B. boiling point, C. subilmation
bulgar [2K]
 I am pretty sure that <span>the following whihc cannot be determined by looking at the phase diagram is definitely </span>D. system pressure.  I consider this one to be correct because only this point is not included into<span> phase diagram and can't be determined itself. Hope it will help! Regards!</span>
7 0
3 years ago
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