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Paul [167]
3 years ago
6

CHEGG You have two pipes, having the same elevation, connected to each other that transport water. The given density of water is

rho = 997 kg/m . The flow rate of water through the pipes is R = .009463522m3/s, but they have different diameters. The first pipe has a measured pressure P 1=13.785????Pa and a velocity of v = 41.221 m/s. The second pipe has a measured velocity of v = 55.654m/s. A) What is the pressure in the second pipe? Provide your answer in MPa.
Physics
1 answer:
Yakvenalex [24]3 years ago
3 0

Answer:

P = 0.6814 MPa

Explanation:

For this exercise we must use the Betuliano relation

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

where we use the subscript 1 for the first pipe and the subscript 2 for the second pipe

indicate that the two pipes are at the same level therefore

        y₁ = y₂

       P₂ = P₁ + ½ ro (v₁² - v₂²)

let's calculate

       P₂ = 13,785 10⁵ + ½ 997 (41,221² - 55,654²)

       P₂ = (13.785 - 6.97) 105

       P₂ = 6.814 10⁵ Pa

let's reduce to Mpa

       P₂ = 6.814 10⁵ Pa (1 MPa / 10⁶ Pa)

       P₂ = 6.814 10-1 MPa

       P = 0.6814 MPa

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A 2130 kg car is parked on a hill that makes a 15º angle with the horizontal. What is the normal force on the car?
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Answer:

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Explanation:

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F = 2130 x 10 x cos 15⁰

F = 2130 x 10 x 0.9659

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4 years ago
Two planets X and Y travel counterclockwise in circular orbits about a star, as seen in the figure.
sergeinik [125]

Planet Y has rotated by 135.5° through during this time.

To find the answer, we need to know about the relation between angle and radius of orbit.

<h3>What's the expression of angle in terms of radius?</h3>
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  • As arc = orbital velocity × time,

            angle= (orbital velocity × time)/radius

  • Orbital velocity= √(GM/radius), G= gravitational constant and M = mass of sun
  • So, angle = (√(GM)× time)/radius^3/2
<h3>What's is the angle rotated by planet Y after 5 years, if ratio of the radius of orbit of planet X and Y is 4:3 and planet X is rotated by 88°?</h3>
  • Let Ф₁= angle rotated by planet Y, Ф₂= angle rotated by planet X
  • As time = 5 years ( a constant)
  • Ф₁/Ф₂= (radius of planet X / radius of planet Y)^(3/2)
  • Ф₁= (radius of planet X / radius of planet Y)^(3/2) × Ф₂

   = (4/3)^(3/2) × 88°

   = 135.5°

Thus, we can conclude that Planet Y has rotated by 135.5° through during this time.

Learn more about the orbital velocity here:

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8 0
2 years ago
Two men, Joel and Jerry, each pushes an object that are identical on a horizontal frictionless floor starting from rest. Joel an
Nataliya [291]

Answer:

The work done by Joel is greater than the work done by Jerry.

Explanation:

Let suppose that forces are parallel or antiparallel to the direction of motion. Given that Joel and Jerry exert constant forces on the object, the definition of work can be simplified as:

W = F\cdot \Delta s

Where:

W - Work, measured in joules.

F - Force exerted on the object, measured in newtons.

\Delta s - Travelled distance by the object, measured in meters.

During the first 10 minutes, the net work exerted on the object is zero. That is:

W_{net} = W_{Joel} - W_{Jerry}

W_{net} = F\cdot \Delta s - F\cdot \Delta s

W_{net} = (F-F)\cdot \Delta s

W_{net} = 0\cdot \Delta s

W_{net} = 0\,J

In exchange, the net work in the next 5 minutes is the work done by Joel on the object:

W_{net} = W_{Joel}

W_{net} = F\cdot \Delta s

Hence, the work done by Joel is greater than the work done by Jerry.

7 0
4 years ago
Two wires are made of the same material. Wire 1 has length that is 1.35 times the length of wire 2 and diameter that is 0.91 tim
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Answer:

1.117935:1

Explanation:

Since the wires are of the same material, they will have the same resistivity \rho.

The cross-sectional area of the of a wire is given by;

A=\pi\frac{d^2}{4}................(1)

where d is the diameter of the wire.

Also, the relationship between resistance R, cross-sectional area A and length l of a wire is given as follows;

\rho=\frac{RA}{l}..................(2)

Since the resistivity same for both wires, say wire 1 and wire 2, we can wreite the following;

\frac{R_1A_1}{l_1}=\frac{R_2A_2}{l_2}..................(3)

Hence from eqaution (3), the ration of wire 1 to 2 is expressed as;

\frac{R_1}{R_2}=\frac{l_1A_2}{l_2A_1}..................(4)

Given;

l_1=1.35l_2

\frac{R_1}{R_2}=\frac{1.35l_2A_2}{l_2A_1}\\\frac{R_1}{R_2}=\frac{1.35A_2}{A_1}.............(5)

We then use equation (1) to fine the ratio of the area A_1 to A_2

bearing in mind that d_1=0.91d_2

This ratio gives 0.8281. Substituting this into equation (5), we get the following;

\frac{R_1}{R_2}= 1.35*0.8281=1.117935

4 0
3 years ago
Read 2 more answers
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