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Alex
3 years ago
13

If you see a lightning stroke and then, 15 seconds later, hear the thunder, the lightning is about ____ miles away

Physics
1 answer:
makvit [3.9K]3 years ago
3 0

Answer:

3.2 miles

Explanation:

We can assume that the light coming from the lightning reach instantaneously our eyes (because light travels very fast, 3\cdot 10^8 m/s), so the time between the lightining and the hearing of the thunder is equal to the time that sound takes to each us.

Sound travels in air approximately at speed

v = 343 m/s

So the distance it covers in a time of t = 15 seconds is

d=vt=(3403 m/s)(15 s)=5145 m

And using the conversion from meters to miles,

1 mile = 1609 meters

we find

d=\frac{5145 m}{1609 m/miles} = 3.2 miles

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8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
Which structure is found in all EUKARYOTIC cells? Large central vacuole, Golgi apparatus, flagella, cilia
Pavel [41]
The answer is Golgi apparatus.
HOPE THIS HELPS!
3 0
3 years ago
Which person is weightless?
ahrayia [7]

Answer: Option (b) is the correct answer.

Explanation:

The force of gravity acting on an object helps in determining the weight of an object. But a place where there will be no gravity or have zero gravitational pull then it means the person will be weightless.

For example, force of gravity on moon is zero which means any object or person on moon will be weightless.

On the other hand, when a child is in the air as she plays on a trampoline then it means gravitational pull form the earth is acting on it. So, it will definitely has some weight.

Similarly, a scuba diver exploring a deep-sea wreck is under the ground where there will be force of gravity. Hence, it will also have some weight.

Thus, we can conclude that an astronaut on the Moon is the person who is weightless.

7 0
3 years ago
A box experiencing a gravitational force of 600 N. is being pulled to the right with a force of 250 NA 25 N. frictional force ac
Hatshy [7]

Answer:

32

Explanation:

3 0
3 years ago
An emergency relief plane is dropping a care package from a plane to a group of medical personnel working for a relief agency in
ipn [44]

Answer:

The altitude of the plane is 379.5 m.

Explanation:

Initial horizontal velocity, u = 59.1 m/s

Horizontal distance, d = 521 m

let the time taken by the packet to cover the distance is t.

Horizontal distance = horizontal velocity x time

521 = 59.1 x t

t = 8.8 s

let the vertical height is h .

Use second equation of motion in vertical direction.

h = u t  + 0.5 gt^2\\\\h = 0 + 4.9 \times 8.8\times8.8\\\\h= 379.5 m

7 0
2 years ago
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