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4 years ago

11

If you could travel to a star 25 light-years away and return to earth at nearly the speed of light, how much time would elapse o

n earth?

2 answers:

anastassius [24]4 years ago

7 0

Millions and millions of years

lyudmila [28]4 years ago

6 0

50 years, would it not be, because time moves on no matter what, so 25ish plus 25ish would be 50 years. Good question

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What is the electric field due to a point charge of 20uC at a distance of 1 meter away from it?

Anettt [7]

The electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 $\frac{N}{C}$ .

First, you have to know that the space surrounding a load suffers some kind of disturbance, since a load located in that space will suffer a force. The disturbance that this charge creates around it is called an electric field.

In other words, an electric field exists in a certain region of space if, when introducing a charge called witness charge or test charge, it undergoes the action of an electric force.

The electric field E created by the point charge q at any point P, located at a distance r, is defined as:

$E=K\frac{q}{r^{2} }$

where K is the constant of Coulomb's law.

In this case, you know:

K= 9×10⁹ $\frac{Nm^{2} }{C^{2} }$
q= 20 uC=20×10⁻⁶ C
r= 1 m

Replacing in the definition of electric field:

$E=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{20x10^{-6} C}{(1 m)^{2} }$

Solving:

<u><em>E=180000 </em></u> $\frac{N}{C}$ <u><em /></u>

Finally, the electric field due to a point charge of 20uC at a distance of 1 meter away from it is 180000 $\frac{N}{C}$ .

Learn more:

5 0

2 years ago

How do machines increase mechanical advantage?

Kryger [21]

Mechanical advantage is a measure of the force amplification achieved by using a tool.

7 0

3 years ago

Can anyone help me on this thnxs

DIA [1.3K]

Answer:

a sin Ø - b cos Ø

_______________. =

a sin Ø+ b cos Ø

=> a - b cot ∅

_____________

a + b cot ∅

=> a - b x b

__

a

______________

a + b x b

___

a

=> a² - b ²

__________

a ² + b²

8 0

3 years ago

A pressure vessel that has a volume of 10m3 is used to store high-pressure air for operating a supersonic wind tunnel. If the ai

Anna71 [15]

Answer:

The mass of air stored in the vessel is 235.34 kilograms.

Explanation:

Let supossed that air inside pressure vessel is an ideal gas, The density of the air ( $\rho$ ), measured in kilograms per cubic meter, is defined by following equation:

$\rho = \frac{P\cdot M}{R_{u}\cdot T}$ (1)

Where:

$P$ - Pressure, measured in kilopascals.

$M$ - Molar mass, measured in kilomoles per kilogram.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meters per kilomole-Kelvin.

$T$ - Temperature, measured in Kelvin.

If we know that $P = 2026.5\,kPa$ , $M = 28.965\,\frac{kg}{kmol}$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}$ and $T = 300\,K$ , then the density of air is:

$\rho = \frac{(2026.5\,kPa)\cdot \left(28.965\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (300\,K)}$

$\rho = 23.534\,\frac{kg}{m^{3}}$

The mass of air stored in the vessel is derived from definition of density. That is:

$m = \rho \cdot V$ (2)

Where $m$ is the mass, measured in kilograms.

If we know that $\rho = 23.534\,\frac{kg}{m^{3}}$ and $V = 10\,m^{3}$ , then the mass of air stored in the vessel is:

$m = \left(23.534\,\frac{kg}{m^{3}} \right)\cdot (10\,m^{3})$

$m = 235.34\,kg$

The mass of air stored in the vessel is 235.34 kilograms.

7 0

3 years ago

A 60 kg student is standing motionless. Approximately how much force is the floor exerting on the student?

Solnce55 [7]

The normal force is equal in magnitude and opposite in direction to the weight of the student.
W = m · g = 60 kg · 10 m/s² = 600 N ( downward )
N = - 600 N
Answer:
B ) 600 N upward.

7 0

3 years ago