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Arturiano [62]
3 years ago
5

Water (rhoH20 = 1000.0 kg/m3 ) flows through a garden hose that goes up a step 20.0 cm high. The cross-sectional area of the hos

e on the top of the step is half that at the bottom of the step. The speed of the water at the bottom of the step is 1.20 m/s. If the water pressure is 143 kPa at the bottom of the step, what is the water pressure at the top of the step?
Physics
1 answer:
Soloha48 [4]3 years ago
3 0

Answer:

 P₂ = 138.88 10³ Pa

Explanation:

This is a problem of fluid mechanics, we must use the continuity and Bernoulli equation

Let's start by looking for the top speed

        Q = A₁ v₁ = A₂ v₂

We will use index 1 for the lower part and index 2 for the upper part, let's look for the speed in the upper part (v2)

         v₂ = A₁ / A₂ v₁

They indicate that A₂ = ½ A₁ and give the speed at the bottom (v₁ = 1.20 m/s)

         v₂ = 2  1.20

         v₂ = 2.40 m / s

Now let's write the Bernoulli equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂

Let's clear the pressure at point 2

       P₂ = P₁ + ½ ρ (v₁² - v₂²) + ρ g (y₁-y₂)

we put our reference system at the lowest point

        y₁ - y₂ = -20 cm

Let's calculate

       P₂ = 143 10³ + ½ 1000 (1.20² - 2.40²) + 1000 9.8 (-0.200)

       P₂ = 143 103 - 2,160 103 - 1,960 103

       P₂ = 138.88 10³ Pa

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A boy is trying to roll a bowling ball up a hill, as shown in the image below. If
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The minimum initial velocity that the ball must have for it to reach the top of the hill is 21 m/s. The correct option is D.

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

A boy is trying to roll a bowling ball up a hill. The friction is ignored.  The ball must have to reach the top of the hill with a velocity. The acceleration due to gravity, g = 9.8 m/s²

The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.

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A metal cylinder with a mass of 4.20 kg is attached to a spring and is able to oscillate horizontally with negligible friction.
kherson [118]

Answer:

a) k = 120 N / m

, b)    f = 0.851 Hz

, c)  v = 1,069 m / s

, d)  x = 0

, e)  a = 5.71 m / s²

, f)   x = 0.200 m

, g)  Em = 2.4 J

, h) v = -1.01 m / s

Explanation:

a) Hooke's law is

         F = k x

         k = F / x

          k = 24.0 / 0.200

          k = 120 N / m

b) the angular velocity of the simple harmonic movement is

        w = √ k / m

        w = √ (120 / 4.2)

        w = 5,345 rad / s

Angular velocity and frequency are related.

       w = 2π f

        f = w / 2π

        f = 5.345 / 2π

        f = 0.851 Hz

c) the equation that describes the movement is

        x = A cos (wt + Ф)

As the body is released without initial velocity, Ф = 0

        x = 0.2 cos wt

Speed ​​is

       v = dx / dt

       v = -A w sin wt

The speed is maximum for sin wt = ±1

       v = A w

       v = 0.200 5.345

       v = 1,069 m / s

d) when the function sin wt = -1 the function cos wt = 0, whereby the position for maximum speed is

       x = A cos wt = 0

       x = 0

e) the acceleration is

       a = d²x / dt² = dv / dt

       a = - Aw² cos wt

The acceleration is maximum when cos wt = ± 1

       a = A w²

        a = 0.2   5.345

        a = 5.71 m / s²

f) the position for this acceleration is

       x = A cos wt

       x = A

       x = 0.200 m

g) Mechanical energy is

        Em = ½ k A²

        Em = ½ 120 0.2²

       Em = 2.4 J

h) the position is

         x = 1/3 A

Let's calculate the time to reach this point

         x = A cos wt

        1/3 A = A cos 5.345t

         t = 1 / w cos⁻¹(1/3)

The angles are in radians

t = 1.23 / 5,345

t = 0.2301 s

Speed ​​is

v = -A w sin wt

v = -0.2 5.345 sin (5.345 0.2301)

v = -1.01 m / s

i) acceleration

a = -A w² sin wt

a = - 0.2 5.345² cos (5.345 0.2301)

      a = -1.91 m / s²

5 0
3 years ago
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