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vampirchik [111]

3 years ago

12

A car moving at a speed of 20 m/s then accelerates uniformly at 1.8 m/s^2 until it reaches a speed of 25 m/s. What distance does

it travel during the time it is accelerating?
66 m
54 m
62 m
50 m
58 m

1 answer:

Volgvan3 years ago

8 0

Answer:

62.5 m

Explanation:

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A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the

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To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

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Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

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We know as well by definition that

$1gal = 3.785*10^{-3}m^3$

Then we have for the statement that

$V_f = V_0 -1gal$

$V_f = V_0 - 3.785*10^{-3}$

Replacing the previous data

$\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}$

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Solving to get h,

$h = 1.99963m$

Therefore the change is

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$\Delta h = 2- 1.99963$

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Therefore te change in the height of the water in the tank is 0.37mm

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3 years ago