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Anna11 [10]
3 years ago
12

Select an example of a vertical motion with a positive velocity and a negative acceleration. Select the correct example and expl

anation.a)A tossed ball on its way down. As the ball moves down it is slowing down, hence the acceleration is opposite the bvelocity (negative).b) A tossed ball on its way upward. As the ball moves upward it is slowing down, hence the acceleration is pointing in the same direction as the velocity (negative).c) A tossed ball on its way upward. As the ball moves upward it is slowing down, hence the acceleration is opposite the velocity (negative).d) A tossed ball on its way down. As the ball moves down it is slowing down, hence the acceleration is pointing in the same direction as the velocity (negative).
Physics
1 answer:
miss Akunina [59]3 years ago
8 0

Answer:

c)

Explanation:

When a ball is tossed upward, it rises due to the initial velocity which points upward (so it is positive if we choose the upward direction as positive), but once released, the only force (neglecting air resistance) acting on the ball is gravity.

This force is due to the mass of the Earth, and produces an acceleration in the same direction than the force, which is always downward, so in this case is negative.

As it has opposite direction to the velocity, causes the ball to slow down until it reach to a point when momentarily comes to an stop, and then changes direction falling with negative velocity, so it speeds up as it has the same sign than acceleration, that it is always present.

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Answer:

D ≈ 8.45 m

L ≈ 100.02 m

Explanation:

Given

Q = 350 m³/s (volumetric water flow rate passing through the stretch of channel, maximum capacity of the aqueduct)

y₁ - y₂ = h = 2.00 m (the height difference from the upper to the lower channels)

x = 100.00 m (distance between the upper and the lower channels)

We assume that:

  • the upper and the lower channels are at the same pressure (the atmospheric pressure).
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  • ρ = 1000 Kg/m³ (density of water)

We apply Bernoulli's equation as follows between the point 1 (the upper channel) and the point 2 (the lower channel):

P₁ + (ρ*v₁²/2) + ρ*g*y₁ = P₂ + (ρ*v₂²/2) + ρ*g*y₂

Plugging the known values into the equation and simplifying we get

Patm + (1000 Kg/m³*(0 m/s)²/2) + (1000 Kg/m³)*(9.81 m/s²)*(2 m) = Patm + (1000 Kg/m³*v₂²/2) + (1000 Kg/m³)*(9.81 m/s²)*(0 m)

⇒ v₂ = 6.264 m/s

then we apply the formula

Q = v*A  ⇒   A = Q/v ⇒   A = Q/v₂

⇒   A = (350 m³/s)/(6.264 m/s)

⇒   A = 55.873 m²

then, we get the diameter of the pipe as follows

A = π*D²/4   ⇒   D = 2*√(A/π)

⇒   D = 2*√(55.873 m²/π)

⇒   D = 8.434 m ≈ 8.45 m

Now, the length of the pipe can be obtained as follows

L² = x² + h²

⇒ L² = (100.00 m)² + (2.00 m)²

⇒ L ≈ 100.02 m

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