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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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<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

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