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ki77a [65]
3 years ago
11

lightning strikes in the distance and six seconds later Thunder is heard how far away was the lightning strike

Physics
1 answer:
marshall27 [118]3 years ago
8 0
The speed of sound really depends on the temperature and moisture in the air,
and the sound of the thunder doesn't necessarily travel straight from the lightning
to where you are.  So we can't say exactly. 

But if we use the nominal speed of 340 m/s, then 6 seconds means 2,040 meters,
or about 6,700 feet, or about 1.27 miles.
You might be interested in
A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When
11111nata11111 [884]

Answer:

The value is R_f =  \frac{4}{5}  R

Explanation:

From the question we are told that

   The  initial velocity of the  proton is v_o

    At a distance R from the nucleus the velocity is  v_1 =  \frac{1}{2}  v_o

    The  velocity considered is  v_2 =  \frac{1}{4}  v_o

Generally considering from initial position to a position of  distance R  from the nucleus

 Generally from the law of energy conservation we have that  

       \Delta  K  =  \Delta P

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

      \Delta K  =  K__{R}} -  K_i

=>    \Delta K  =  \frac{1}{2}  *  m  *  v_1^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * (\frac{1}{2} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

          \Delta P =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P =  k  *  \frac{q_1 * q_2 }{R}  - 0

So

           \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R}  - 0

=>        \frac{1}{2}  *  m  *v_0^2 [ \frac{1}{4} -1 ]  =   k  *  \frac{q_1 * q_2 }{R}

=>        - \frac{3}{8}  *  m  *v_0^2  =   k  *  \frac{q_1 * q_2 }{R} ---(1 )

Generally considering from initial position to a position of  distance R_f  from the nucleus

Here R_f represented the distance of the proton from the nucleus where the velocity is  \frac{1}{4} v_o

     Generally from the law of energy conservation we have that  

       \Delta  K_f  =  \Delta P_f

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus  , this is mathematically represented as

      \Delta K_f   =  K_f -  K_i

=>    \Delta K_f  =  \frac{1}{2}  *  m  *  v_2^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * (\frac{1}{4} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * \frac{1}{16} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R_f  from the nucleus , this is mathematically represented as

          \Delta P_f  =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P_f  =  k  *  \frac{q_1 * q_2 }{R_f }  - 0      

So

          \frac{1}{2}  *  m  * \frac{1}{8} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f }

=>        \frac{1}{2}  *  m  *v_o^2 [-\frac{15}{16} ]  =   k  *  \frac{q_1 * q_2 }{R_f }

=>        - \frac{15}{32}  *  m  *v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f } ---(2)

Divide equation 2  by equation 1

              \frac{- \frac{15}{32}  *  m  *v_o^2 }{- \frac{3}{8}  *  m  *v_0^2  } }   =  \frac{k  *  \frac{q_1 * q_2 }{R_f } }{k  *  \frac{q_1 * q_2 }{R } }}

=>           -\frac{15}{32 } *  -\frac{8}{3}   =  \frac{R}{R_f}

=>           \frac{5}{4}  =  \frac{R}{R_f}

=>             R_f =  \frac{4}{5}  R

   

7 0
3 years ago
In the water cycle matter moves towards the pull of gravity during ??
Bumek [7]

infiltration and evaporation

7 0
3 years ago
The core of a 400 Hz aircraft transformer has a net cross-sectional area of 13 cm2. The maximum flux density is 0.9 T, and there
jenyasd209 [6]

Answer:

32.76 Volt

Explanation:

frequency, f = 400 Hz

Area of crossection, A = 13 cm²

Maximum flux density, B = 0.9 tesla

Number of turns in secondary coil, N = 70

Let the maximum induced voltage is e.

According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.

e = dФ/dt

e=\frac{NBA}{t}

Time is defined as the reciprocal of frequency.

So, e = N B A f

e = 70 x 0.9 x 13 x 10^-4 x 400

e = 32.76 volt

4 0
3 years ago
Thermal expansion results in
PIT_PIT [208]
Thermal expansion<span> is the tendency of matter to change in shape, area, and volume in response to a change in temperature, through </span>heat<span> transfer. Temperature is a monotonic function of the average molecular kinetic energy of a substance. When a substance is heated, the kinetic energy of its molecules increases.so this can result in.......heat</span>
4 0
3 years ago
Read 2 more answers
The radius of the aorta is about 1.4cm , and the blood passing through it has a speed of about 40cm/s .
Lapatulllka [165]

For the average speed of blood flow in the major arteries of the body  is mathematically given as

v2 = 117.29m/s

<h3>What is the average speed of blood flow in the major arteries of the body?</h3>

Generally, the equation for the average speed  is mathematically given as

A1 v1 = A2 v2

(pi r1^2) v1 = A2 v2

(3.14x(1.4)^2 )x 40 = (2.1) xV2

v2 = 117.29m/s

In conclusion, the average speed of blood flow

v2 = 117.29m/s

Read more about Speed

brainly.com/question/4931057

4 0
2 years ago
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