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3 years ago

What are the four factors that influence how fast the water will flow

1 answer:

8 0

<span>Viscosity is a property of the water which opposes the relative motion between the two surfaces of the water that are moving at different velocities. In simple terms, viscosity of water means friction between the molecules of water. Factors affecting viscosity of water
1. TemperatureTemperature strongly influences a fluid’s viscosity. Viscosity decreases with increasing temperature and vice versa.
2. PressureNormally, an increase in pressure causes viscosity to increase.
3. ConcentrationConcentration has also direct relation with the viscosity.<span>
</span></span>

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What crisis is occurring in California?

nlexa [21]

Hey there!

Option A

A climate crisis is occurring in California where sudden forest fires occur due to this .

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2 years ago

Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.

alexandr1967 [171]

Answer:

$v=\sqrt{k\frac{e^2}{m_e r}}$ , $2.18\cdot 10^6 m/s$

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

$k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}$

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

$m_e$ is the mass of the electron

Solving for v, we find

$v=\sqrt{k\frac{e^2}{m_e r}}$

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

$r=5.3\cdot 10^{-11}m$

while the electron mass is

$m_e = 9.11\cdot 10^{-31}kg$

and the charge is

$e=1.6\cdot 10^{-19} C$

Substituting into the formula, we find

$v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s$

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3 years ago

A narrow beam of light from a laser travels through air (n = 1.00) and strikes the surface of the water (n = 1.33) in a lake at

Natalka [10]

Answer:

A) d = 11.8m

B) d = 4.293 m

Explanation:

A) We are told that the angle of incidence;θ_i = 70°.

Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;

tan 70° = d/4.3m

Where d is the distance from point B at which the laser beam would strike the lakebottom.

So,d = 4.3*tan70

d = 11.8m

B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)

So,

n1*sinθ_i = n2*sinθ_r

Thus; sinθ_r = (n1*sinθ_i)/n2

sinθ_r = (1 * sin70)/1.33

sinθ_r = 0.7065

θ_r = sin^(-1)0.7065

θ_r = 44.95°

Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;

d = 4.3 tan44.95

d = 4.293 m

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3 years ago

Why is it a good idea to explore the health and fitness field in terms of career

KIM [24]

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3 years ago

Read 2 more answers

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

Zepler [3.9K]

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, $a=3\ m/s^2$

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

$m=\dfrac{W}{g}$

$m=\dfrac{150}{9.8}$

$m=15.3\ kg$

Frictional force is given by :

$F=ma$

$F=15.3\times 3$

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

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3 years ago