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Archy [21]
2 years ago
5

What is the speed in meters per second of a car that is travelling at 82km/h?

Physics
2 answers:
4vir4ik [10]2 years ago
5 0
Divide by 3.6
82/3.6 = 22.8 m/s
suter [353]2 years ago
3 0

Answer:

22.777777777777 m/s

Explanation:

Since 1 kilometer is 1000 meters, then 82 km is 82000 meters. Since one hour is the same as 3600 seconds, then the answer is pretty simple: 82000/3600 or 820/36. That gets you 22.77777 or 22.78. Hope this helps

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You could use grams hope this helps
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Which of the following is a term for the process of rock being broken down into pieces by wind or moving water?
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2 years ago
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At an amusement park, a swimmer uses a water slide to enter the main pool. a. If the swimmer starts at rest, slides with negligi
adoni [48]

Answer:

a)6.7m/S

b)6.8m/s

Explanation:

Hello ! To solve the point b you must follow the steps below

1.Draw the slide taking into account its length and height and find the angle from which the swimmer is launched (see attached image)

2. Find the horizontal velocity (X) and vertical (Y) components (see attached image)

3) for the third step we must remember that as in the slide there is no horizontal acceleration the speed in X will remain constant at the end of the swimmer's path (Vx = 0.59m / s)

4)

the fourth step is to remember that vertically there is constant acceleration called gravity (g = 9.81m / s ^ 2), so to find the speed at the end of the route we use the following equation

Vfy= \sqrt{Vy^2+2gy}

where    

Vfy= final verticaly speed    

Vy=initial verticaly speed=0.59m/S

g=gravity=9.81m/S^2

y=height of slide=2.31m

solving

Vfy= \sqrt{Vy^2+2gy}\\Vfy= \sqrt{(0.59)^2+2(9.81)(2.31)}=6.77m/s

The last step is to add the velocity components vectorally at the end of the route with the following equation

V=\sqrt{Vfy^2+Vx^2} =\sqrt{6.77^2+0.59^2} =6.8m/s

point A

taking into account the previous steps we can infer that as the swimmer starts from rest, the velocity (Vx=Vy=O) is zero, so we should only use the formula for constant acceleration movement.

Vfy= \sqrt{Vy^2+2gy}

vy=0

Vfy=\sqrt{2gy}

Vfy=\sqrt{2(9.81)(2.31)}=6.7m/s

8 0
3 years ago
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dem82 [27]
Static friction

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Xplain the difference between aerobic and anaerobic exercise and give 1 example of each?
Lilit [14]

Answer:

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Explanation:

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