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4 years ago

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El número de Reynolds para un fluido de densidad y viscosidad y que se mueve con velocidad media v a través de un conducto cualq

uiera, está definido como:
NR=vDn

Demuestre que si el fluido se mueve por un tubo de sección transversal circular de diámetro D, el número de Reynolds para este caso se puede expresar como:
NR=4JπηD
Donde J=mt es el flujo de masa.

1 answer:

choli [55]4 years ago

4 0

Answer:

reeeee

Explanation:

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A new planet has been discovered that has a mass one-sixth that of Earth and a radius that is six times that of Earth. Determine

Soloha48 [4]

Answer:

0.045 m/s²

Explanation:

Let the mass of Earth be 'M' and radius be 'R'.

Given:

Mass of the new planet (m) = one-sixth of Earth's mass = $\frac{M}{6}$

Radius of new planet (r) = 6 times Earth's radius = $6R$

We know that, acceleration due to gravity of a planet of mass 'M' and radius 'R' is given as:

$g=\dfrac{GM}{R^2}$

Now, this is acceleration due to gravity on Earth.

Now, acceleration due to gravity of new planet is given as:

$g_{new}=\dfrac{Gm}{r^2}\\\\g_{new}=\dfrac{G\times\frac{M}{6}}{(6R)^2}\\\\g_{new}=\dfrac{GM}{6\times 36R^2}\\\\g_{new}=\frac{1}{216}(\frac{GM}{R^2})=\frac{1}{216}\times g
$

Now, the value of 'g' on Earth is approximately 9.8 m/s². So,

$g_{new}=\frac{9.8}{216}=0.045\ m/s^2$

Therefore, the free fall acceleration on the surface of this planet is 0.045 m/s².

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4 years ago

Which would likely demagnetize a magnetized iron nail?

Delicious77 [7]

Dropping it on a hard surface.

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3 years ago

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Why is it important to understand size, function and structure of the universe?

SpyIntel [72]

How about to understand the nature of reality itself? Also, we live in the universe, so it might be useful to understand how it works.

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4 years ago

A ball is thrown vertically upward from the top of a building 80 feet tall with an initial velocity of 64 feet per second. The d

-BARSIC- [3]

Answer:

a) $t=6.37s$

b) $t=3.3333s$

Explanation:

The knowable variables are the initial hight and initial velocity

$s_{o}=80ft$

$v_{os}=64ft/s$

The equation that describes the motion of the ball is:

$s=80+64t-16t^{2}$

If we want to know the time that takes the ball to hit the ground, we need to calculate it by doing s=0 that is the final hight.

$0=80+64t-12t^{2}$

a) Solving for t, we are going to have two answers

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

a=-16

b=64

c=80

$t=-1.045 s$ or $t=6.378s$

<em><u>Since time can not be negative the answer is t=6.378s </u></em>

b) To find the time that takes the ball to pass the top of the building on its way down, we must find how much does it move too

First of all, we need to find the maximum hight and how much time does it take to reach it:

$v_{y}=v_{o}+gt$

at maximum point the velocity is 0

$0=64-32.2t$

Solving for t

$t=1.9875 s$

Now, we must know how much distance does it take to reach maximum point

$s=0+64t-16t^{2} =64(1.9875)-12(1.9875)^{2} =80ft$

So, the ball pass the top of the building on its way down at 160 ft

$160=80+64t-16t^{2}$

Solving for t

$t=2s$ or $t=3.333s$

Since the time that the ball reaches maximum point is almost t=2s that answer can not be possible, so the answer is t=3.333s for the ball to go up and down, passing the top of the building

4 0

3 years ago

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Magnitude of force F experienced by a certain object moving with speed v is given by F=Kv race to 2, where K is constant. Find t

My name is Ann [436]

<h2>Answer:</h2>

MT⁻¹

<h2>Explanation:</h2>

Given equation:

F = Kv --------------(i)

Where;

F = Magnitude of force experienced by the object

v = speed at which the object is moving

K = constant.

To get the dimension of K, follow the following steps:

i. <em>Make K subject of the formula in equation(i)</em>

K = F / v

ii.<em> Get the dimension of force F and speed v</em>

Dimension of force = MLT⁻²

Dimension of velocity = LT⁻¹

iii.<em> Substitute these dimensions into the result of (i) above;</em>

K = MLT⁻² / LT⁻¹

iv. <em>Simplify the result in (iii)</em>

<em>First, the </em><em>L</em><em>s in the numerator and denominator will cancel out</em>

K = MT⁻² / T⁻¹

<em>Next, the </em><em>T</em><em>s will be expressed as follows</em>

K = MT⁻² x T¹ [using the law of indices]

K = MT⁻²⁺¹

K = MT⁻¹

Therefore, the dimension of K is MT⁻¹

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3 years ago