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alukav5142 [94]
3 years ago
6

Gasoline flows in a long, underground pipeline at a constant temperature of 15o C (rho = 680 kg/m3 ; ν = 4.6 × 10-7 m2 /s). Two

pumping stations are located at the same elevation 13 km apart. The pressure drop between the stations is 1.4 MPa. The diameter of the pipe is 0.6 m and has a roughness of ε = 0.12 mm. Determine the volume flow rate.

Physics
1 answer:
poizon [28]3 years ago
8 0

Answer:

1.0416 m∧3/sec

Explanation:

check the pictures below for the solution

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A loop of wire in the shape of a rectangle rotates with a frequency of 219 rotation per minute in an applied magnetic field of m
bazaltina [42]

Answer:

Emax = 0.055V

Imax = 7.86mA

Explanation:

See attachment below.

6 0
3 years ago
Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant
irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

a_g = \frac{GM}{R^2}

Here

M = \text{Mass inside the Orbit of the star}

R = \text{Orbital radius}

G = \text{Universal Gravitational Constant}

Mass inside the orbit in terms of Volume and Density is

M =V \rho

Where,

V = Volume

\rho =Density

Now considering the volume of the star as a Sphere we have

V = \frac{4}{3} \pi R^3

Replacing at the previous equation we have,

M = (\frac{4}{3}\pi R^3)\rho

Now replacing the mass at the gravitational acceleration formula we have that

a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho

a_g = \frac{4}{3} G\pi R\rho

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration.  So centripetal acceleration of the star is

a_c = \frac{4}{3} G\pi R\rho

At the same time the general expression for the centripetal acceleration is

a_c = \frac{\Theta^2}{R}

Where \Theta is the orbital velocity

Using this expression in the left hand side of the equation we have that

\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2

\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}

\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R

Considering the constant values we have that

\Theta = \text{Constant} \times R

\Theta \propto R

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0
3 years ago
The figure below shows two forces A=15 N and B=5 N acting on an object. What is the
alexandr1967 [171]

Answer:

18.03 N

Explanation:

From the fiqure below,

Using parallelogram law of vector

R² = 15²+5²-2×5×15cos(180-60)

R² = 225+25-150cos120°

R² = 250-150(-0.5)

R² = 250+75

R² = 325

R = √325

R = 18.03 N

Hence the resultant force of the object is 18.03 N

7 0
3 years ago
2. A bus drove 8 meters East, then turned to drive 8 meters North, then 2 meters
Sphinxa [80]

Answer:

Long question good luck:)

Explanation:

3 0
2 years ago
For sprinters running at 12 m/s around a curved track of radius 26 m, how much greater (as a percentage) is the average total fo
Snezhnost [94]

Answer:

114.86%

Explanation:

In both cases, there is a vertical force equal to the sprinter's weight:

Fy = mg

When running in a circle, there is an additional centripetal force:

Fx = mv²/r

The net force is found with Pythagorean theorem:

F² = Fx² + Fy²

F² = (mv²/r)² + (mg)²

F² = m² ((v²/r)² + g²)

F = m √((v²/r)² + g²)

Compared to just the vertical force:

F / Fy

m √((v²/r)² + g²) / mg

√((v²/r)² + g²) / g

Given v = 12 m/s, r = 26 m, and g = 9.8 m/s²:

√((12²/26)² + 9.8²) / 9.8

1.1486

The force is about 114.86% greater (round as needed).

5 0
3 years ago
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