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Lapatulllka [165]
3 years ago
12

If astronauts could travel at v disagree. 0.945c, we on Earth would say it takes 4.200 945 4 44 years to reach pha centaur 42Ф

Çgint ears away. The astronauts (a) How much time passes on the astronauts' clocks? Your response is within 10% of the correct value. This may be due to roundoff error or you could have a mistake in your calculation. results to at least four-digit accuracy to minimize roundoff error. years arn out all men mediate (b) What is the distance to Alpha Centauri as measured by the astronauts?
Physics
1 answer:
kolezko [41]3 years ago
6 0

Answer:

1.3734 years pass on the astronauts' clock.

The distance measured by astronauts is 1.22*10^{16} \ m

Explanation:

The fomula for time dilation is:

t'=t\sqrt{1-\frac{v^2}{c^2} }

where t' is the time observed by the astronauts, t is the time observed form Earth, v is the velocity of the astronaut and c is the velocity of light in a vacuum.

Replacing our values: t=4.2 year, v=0.945c

t'=4.2\sqrt{1-\frac{{(0.945c)}^2}{c^2} } =\\\\=4.2\sqrt{1-0.8930 }=4.2*0.3270=1.3734\ years

The distance measured by astronauts will be their speed multolied by the time is takes them to get to alpha centauri:

D=v*t'=0.945*c*1.3734=0.945*3.00*10^8 m/s*1.3734*3.154*10^7

To acomodate fot units we write the speed of light in m/s and the amount of seconds in a year, so the resulting distance will be in meters.D=1.22*10^{16} \ m

We should compare this to the distance measured from Earth which is D=c*t=3.00*10^8 m/s*4.2*3.154*10^7=3.974*10^{16}\ m

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Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
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The distance is 1.69 m.

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We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

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x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

5 0
3 years ago
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