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musickatia [10]
3 years ago
10

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved

perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 9 foot radius. It is 8 feet tall and has 7.5 feet of water in it. How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the density of the Kool-Aid contaminated water is σ=63.5lbs/ft3
Physics
2 answers:
oksian1 [2.3K]3 years ago
3 0

Answer:

Explanation:

The volume of contaminated water

= cross sectional area x height of water level

3.14 x 9 x 9 x 7.5 ft³

= 1907.55 ft³

mass = density x volume

= 1907.55 x 63.5 lbs

m = 121129.425 lbs

This mass has to be raised to the height of 8 ft before evacuation .

There is a rise of centre of mass of

8 - 7.5/2 ft

h = 4.25 ft

Energy required

= mgh

= 121129.425 x 32 x 4.25

= 16473601.8 unit.

skelet666 [1.2K]3 years ago
3 0

Answer:

Workdone= 8.08×10^3Joules

Explanation:

dV= pir^2 dx

dV = pi×9^2 = 81pi

dF= density× pi× r^2 dx

dF= 63.5 × 81pi dx= 5143.5pi dx

W= Integral from7.5ft of water in the pool(8 -×)dx

W= 5143.5pi(8-7.5)

W= 5143.5 × 3.142 × 0.5

W= 8080.44Joules = 8.08×10^3Joules

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F. If the shuttle's period is synchronized with that of Earth's rotation, what is the height of the shuttle? (1 day = 8.64x104s,
oee [108]

Answer:

1.324 × 10⁷ m

Explanation:

The centripetal acceleration, a at that height above the earth equal the acceleration due to gravity, g' at that height, h.

Let R be the radius of the orbit where R = RE + h, RE = radius of earth = 6.4 × 10⁶ m.

We know a = Rω² and g' = GME/R² where ω = angular speed = 2π/T where T = period of rotation = 1 day = 8.64 × 10⁴s (since the shuttle's period is synchronized with that of the Earth's rotation), G = gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², ME = mass of earth = 6 × 10²⁴ kg. Since a = g', we have

Rω² = GME/R²

R(2π/T)² = GME/R²

R³ = GME(T/2π)²

R = ∛(GME)(T/2π)²

RE + h = ∛(GMET²/4π²)

h = ∛(GMET²/4π²) - RE

substituting the values of the variables, we have

h = ∛(6.67 × 10⁻¹¹ Nm²/kg² × 6 × 10²⁴ kg × (8.64 × 10⁴s)²/4π²) - 6.4 × 10⁶ m

h = ∛(2,987,477 × 10²⁰/4π² Nm²s²/kg) - 6.4 × 10⁶ m

h = ∛75.67 × 10²⁰ m³ - 6.4 × 10⁶ m

h = ∛(7567 × 10¹⁸ m³) - 6.4 × 10⁶ m

h = 19.64 × 10⁶ m - 6.4 × 10⁶ m

h = 13.24 × 10⁶ m

h = 1.324 × 10⁷ m

3 0
3 years ago
Select the correct answer.
Nina [5.8K]

Answer:

Ос.

The spacing between particles increases.

Heat causes greater particle activity, and the particles

are farther apart.

3 0
3 years ago
Objects that rest have no forces upon them
Pani-rosa [81]

Answer:

false

Explanation:

every object will always have the force of gravity acting upon it.

6 0
3 years ago
Water pressurized to 3.5 x 105 Pa is flowing at 5.0 m/s in a horizontal pipe which contracts to 1/3 its former area. What are th
masha68 [24]

Answer:

the pressure after contraction is 2×10^5 Pa

the speed after contraction is 15m/s

Explanation:

We were given Pressure P to be 3.5 x 10^5 that is Flowing with speed of 5.0 m/s,

For us to calculate pressure we need to calculate the area first as;

Let initial Area = A₁

And Final area A₂

We were told that in a horizontal pipe it contracts to 1/3 its former area. Which means

A₂= A₁/3.................

V₁ is the speed

the pressure and speed of the water after the contraction can be calculated using equation of continuity below

A₂V₂ = A₁V₁

But

If we substitute given value in the expresion we have

V₂ = (3A *5)/A

V₂ = 15m/s

Therefore, the speed after contraction is 15m/s

Now we can calculate the pressure using

Bernoulli's equation

p₁ + ½ρv₁² + ρgh₁ = p₂ + ½ρv₂² + ρgh₂

But we know that the pipe is horizontal, then "h" terms cancel out then

p₁ + ½ρv₁² = p₂ + ½ρv₂²

Making P₂ subject of formula we have

p₂ = 0.5ρ( V ₁² - v₂² ) + P₁

P₂=. 0.5 × 1000 (5² -15² ) + 3*10^5

=2×10^5 Pa

Therefore, the pressure after contraction is 2×10^5 Pa

7 0
3 years ago
Why are clouds considered to be a part of the hydrosphere?
blondinia [14]
It is considered parts of the Hydrosphere because water condenses in clouds and falls as precipitation so it is a vital part of the hydrosphere.
7 0
4 years ago
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