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11111nata11111 [884]
3 years ago
12

What must be the pressure difference between the two ends of a 2.6 km section of pipe, 36 cm in diameter, if it is to transport

oil (rho = 950 kg/m^3, η = 0.20 Pa*s) at a rate of 900 cm^3/s ? Express your answer to two significant figures and include the appropriate units. Δ P = 1.3 x 10^3Pa
Physics
1 answer:
ad-work [718]3 years ago
7 0

Answer: 1.13(10)^{3} Pa

Explanation:

This problem can be solved by the following equation:

\Delta P=\frac{8 \eta L Q}{\pi r^{4}}

Where:

\Delta P is the pressure difference between the two ends of the pipe

\eta=0.20 Pa.s is the viscosity of oil

L=2.6 km=2600 m is the length of the pipe

Q=900 \frac{cm^{3}}{s} \frac{1 m^{3}}{(100 cm)^{3}}=0.0009 \frac{m^{3}}{s} is the Rate of flow of the fluid

d=36 cm=0.36 m is the diameter of the pipe

r=\frac{d}{2}=0.18 m is the radius of the pipe

Soving for \Delta P:

\Delta P=\frac{8 (0.20 Pa.s)(2600 m)(0.0009 \frac{m^{3}}{s})}{\pi (0.18 m)^{4}}

Finally:

\Delta P=1135.26 Pa \approx 1.13(10)^{3} Pa

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katrin2010 [14]

Answer:

moderate amount of precipitation

two seasons with temperature differences

Explanation:

Temperate marine climates is also known as oceanic climate and areas who experience this climate are closer to oceans. There is relative precipitation all round the year due to the close proximity to the water body and have the hot and cold season.

The temperate continental climate also has two season and relative rainfall but found in the areas along the Northern hemispheres.

Both climates have moderate amount of precipitation and two seasons with temperature differences.

6 0
3 years ago
A positively charged particle 1 is at the origin of a Cartesian coordinate system, and there are no other charged objects nearby
8090 [49]

Answer:

P=(2 nm, 8mn)

Explanation:

Given :

Position of positively charged particle at origin, O=(0\ nm,0\ nm)

Position of desired magnetic field, D\equiv(1\ nm,8\ nm)

Magnitude of desired magnetic field, E=0\ N.C^{-1}

Let q be the positive charge magnitude placed at origin.

<u>We know the distance between the two Cartesian points is given as:</u>

d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}

<u>For the electric field effect to be zero at point D we need equal and opposite field at the point.</u>

\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 } =\frac{1}{4\pi.\epsilon_0} \times \frac{q}{r^2 }

\therefore (1-0)^2+(8-0)^2=r^2

r^2=65\ nm

r=\sqrt{65}

as we know that the electric field lines emerge radially outward of a positive charge so the second charge will be at equally opposite side of the  given point.

assuming that the second charge is placed at (x,y) nano-meters.

Therefore,

x=2\times 1=2\ nm

and

y=2\times 8=16\ nm

3 0
4 years ago
True or false<br> Lightning happens because of the release of static electricity.
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Yes it is true , lightning happens because of the release of static electricity
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3 years ago
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If the distance between two objects is decreased to one fourth of the original distance, how will this decrease change the force
irinina [24]
<span>D) The new force will be sixteen times greater than the original
This is because of the equation for gravitational force. 
Distance is squared.</span>

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3 years ago
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Is there a point between a 10 nc charge and a 20 nc charge at which the electric field is zero?
KonstantinChe [14]

Let 10 nC and 20 nC charge is placed at "d" distance

now let say we will have a point at distance "x" from 10 nC at which electric field will be zero

now we will have

E_1 = E_2

\frac{kq_1}{x^2} = \frac{kq_2}{(d-x)^2}

now plug in all data

\frac{k(10 nC)}{x^2} = \frac{k(20 nC)}{(d-x)^2}

\frac{10}{x^2} = \frac{20}{(d-x)^2}

square root both sides

\frac{1}{x} = \frac{\sqrt2}{d - x}

d - x = \sqrt2 x

d = x(1 + \sqrt2)

x = \frac{d}{1 + \sqrt2}

so yes there will exist a point between 10 nC and 20 nC charge at which electric field will be zero

3 0
4 years ago
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