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3 years ago

14

How long would it take to drive to alpha centauri (4.4 light-years away)?

1 answer:

oee [108]3 years ago

6 0

A light year is a unit of length that is used to espress distances that are astronomical. It is equivalent to 5.9 trillion miles or 9.5 quadrillion meters. Light year is defined by the distance a beam of light would travel in one year. One light year is equivalent to the distance that light would travel in vacuum on one Julian year or 365 days as defined by the IAU or the International Astronomical Union. So, one light year is traveled for 1 year here on Earth. So, in order to drive to the alpha centauri, you would take about 4.4 years to do so.

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A car enters the freeway with a speed of 6.4 m/s and accelerates uniformly for 3000 m in 210 seconds. How fast is the car moving

dedylja [7]

Answer:

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3 years ago

Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

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Answer:

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Answer:

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