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harkovskaia [24]
2 years ago
14

Johnny was playing baseball with his friends and they noticed a bolt of lightning. They heard thunder seven seconds later. How f

ar away is the storm?
Physics
2 answers:
Thepotemich [5.8K]2 years ago
4 0

That particular strike was very roughly 2.4 km (1.5 miles) away from them.

That's if you use 340 m/s (1120 ft/sec) for the speed of sound. 
But the air in the region for several thousand feet around a thunderstorm
is doing weird things to sounds that pass through it, so you can't use any
exact number for the speed of sound in a stormy area.

The only thing you can be absolutely sure of is that Johnny and his friends
need to round up their equipment and get in the house.  NOW ! 
murzikaleks [220]2 years ago
3 0

2,317 meters on edg!!

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A damped harmonic oscillator consists of a block of mass 2.5 kg attached to a spring with spring constant 10 N/m to which is app
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Answer:

0.5% per oscillation

Explanation:

The term 'damped oscillation' means an oscillation that fades away with time. For Example; a swinging pendulum.

Kinetic energy, KE= 1/2×mv^2-------------------------------------------------------------------------------------------------------------(1).

Where m= Mass, v= velocity.

Also, Elastic potential energy,PE=1/2×kX^2----------------------------------------------------------------------------------------------------------------------(2).

Where k= force constant, X= displacement.

Mechanical energy= potential energy (when a damped oscillator reaches maximum displacement).

Therefore, we use equation (3) to get the resonance frequency,

W^2= k/m--------------------------------------------------------------------------------------(3)

Slotting values into equation (3).

= 10/2.5.

= ✓4.

= 2 s^-1.

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F^2= (-0.1)^2

Potential energy,PE= 1/2 ×0.01

Potential energy= 0.05 ×100

= 0.5% per oscillation.

6 0
1 year ago
Identifying the target population or target audience occurs during which step of the engineering design loop A identified the ne
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1 year ago
Let’s now multiply two numbers in scientific notation using Google.
olga_2 [115]
We are going to rewrite each number:
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5 0
2 years ago
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In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

Distance of charon r_{c} = 19600 \times 10^{3} m

Time period of charon T_{c} = 6.39 days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

   T^{2} = r^{3}

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For first satellites,

  \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} }  }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0
2 years ago
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