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Vinil7 [7]
3 years ago
10

If a storm is 7. 5 kilometers away, how much time is expected between observations of lightning and thunder? Round your answer t

o one decimal place. Seconds.
Physics
1 answer:
Rashid [163]3 years ago
7 0

The expected time period between the observations of lightning and thunder would be 22.5 s.

How to calculate the time of sound waves?

The time taken by sound to reach a certain position can be calculated by the speed formula,

S = \dfrac dt

Where,

S - speed = 0.343 km/s

d - distance = 7.5 km

t - time = ?

Put the values in the formula,

0.343 = \dfrac {7.5} t\\\\t = \dfrac {0.343}{7.5}\\\\t = 22.5

Therefore, the expected time period between the observations of lightning and thunder would be 22.5 s.

Learn more about  the expected time:

brainly.com/question/16851276

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Explanation:

The misunderstanding here is that the thing that turns on the light bulb is not the same electrons near the light switch. So, the electrons near the switch is not moving all the way across the circuit instantly. The electrons are distributed across the wire. When the light switch is turned on, the circuit is connected and there is a potential difference between the bulb and the source. This potential difference creates an electric field, and free electrons move under the influence of this electric field according to Coulomb's Law. When they start to move the electrons closest to the bulb causes the bulb to glow.

So, the important factor here is not the drift velocity of the electrons but the number of electrons and the strength of the electric field.

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3 years ago
a worker uses a board that is 7 m long to pry up a bolder A small rock is used for the fulcrum and is placed 2.5 m from the resi
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4 0
3 years ago
A machine has an efficiency of 80%. How much work must be done on the machine so to make it do 50,000 J of output work?
Tamiku [17]

Work formula is Work=N∙M or Joule (J)

So you have the following given:

50 000 is the output work

.8 is the efficiency

 if you input the given = 50000 = .8*J

To get the answer just divide 50000 by .8 to get the answer.

62,500 J is the amount of work to be done.

8 0
3 years ago
Read 2 more answers
You are planning a trip to the beach 250 miles away. How long will it take if you think you can average 55 miles per hour?
Advocard [28]

Answer: that’s a good question

Explanation:

.

3 0
4 years ago
(III) A baseball is seen to pass upward by a window with a vertical speed of If the ball was thrown by a person 18 m below on th
Ghella [55]

Answer:

<em><u>Assuming that the vertical speed of the ball is 14 m/s</u></em> we found the given values:

a) V₀ = 23.4 m/s

b) h = 27.9 m

c) t = 0.96 s

d) t = 4.8 s

 

Explanation:

a) <u>Assuming that the vertical speed is 14 m/s</u> (founded in the book) the initial speed of the ball can be calculated as follows:  

V_{f}^{2} = V_{0}^{2} - 2gh

<u>Where:</u>

V_{f}: is the final speed = 14 m/s

V_{0}: is the initial speed =?

g: is the gravity = 9.81 m/s²

h: is the height = 18 m

V_{0} = \sqrt{V_{f}^{2} + 2gh} = \sqrt{(14 m/s)^{2} + 2*9.81 m/s^{2}*18 m} = 23.4 m/s  

b) The maximum height is:

V_{f}^{2} = V_{0}^{2} - 2gh

h = \frac{V_{0}^{2}}{2g} = \frac{(23. 4 m/s)^{2}}{2*9.81 m/s^{2}} = 27.9 m

c) The time can be found using the following equation:

V_{f} = V_{0} - gt

t = \frac{V_{0} - V_{f}}{g} = \frac{23.4 m/s - 14 m/s}{9.81 m/s^{2}} = 0.96 s

d) The flight time is given by:

t_{v} = \frac{2V_{0}}{g} = \frac{2*23.4 m/s}{9.81 m/s^{2}} = 4.8 s

         

I hope it helps you!    

3 0
4 years ago
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