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3 years ago

Differences Between light year and astronomical unit in two points .

Physics

2 answers:

Marina CMI [18]3 years ago

7 0

Answer:

A light year is the distance light travels in a year. ... And an astronomical unit is the average distance between the earth and the sun. So the distance to the sun is by definition one AU. A parsec is the distance at which one astronomical unit subtends an angle of one second of arc.

Maslowich3 years ago

5 0

Answer: A light year is the distance light travels in a year. ... And an astronomical unit is the average distance between the earth and the sun. So the distance to the sun is by definition one AU. A parsec is the distance at which one astronomical unit subtends an angle of one second of arc

Explanation:

You might be interested in

If 478 watts of power are used in 14 seconds,how much work was done

zepelin [54]

Answer:

6692J

Explanation:

Power is defined as the rate at which work is being done.

So,

Power = $\frac{workdone}{time }$

Work done = Power x time

Given parameters:

Power = 478watts

Time = 14s

So;

Work done = 478 x 14 = 6692J

6 0

3 years ago

When two capacitors are connected in parallel and then connected to a battery, the total stored energy is 6.9 times greater than

Mumz [18]

Answer:

C1/C2 = 0.213 or C2/C1 = 4.68

Explanation:

Please refer to the attached image for step by step explanation.

5 0

3 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

Lorentz transformation

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

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

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

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

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

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

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

a 1 gram spiders sits on a platform rotating at 78 rpm. the spider is 15 cm from the centre disk. find the speed of the spider

olga nikolaevna [1]

The spider is traveling in a circle with radius = 15cm

The circumference of any circle = 2 pi (radius)
The circumference of the spider's path = 2 pi (15 cm) = 30 pi cm

The spider completes a trip around this path 78 times per minute.
Its speed, relative to you, is

 (78) x (30 pi) cm/min =

 2,340 pi cm/min = 7,351.33 cm/min =

  73.5133 meter/min =

 4.411 km/hr =

 2.74 miles/hour

(After the last appearance of pi,
all numbers are rounded.)

8 0

3 years ago

If you were to drive an average velocity of 65 mi/hr to Traverse City, which is

Amanda [17]

Find the number of hours by dividing the distance by mph. The number of hours will be to the left of the decimal point:

250 miles / 65 mph
= 3.846153846
= 3 hours

2) Find the number of minutes by multiplying what is remaining from step 1 by 60 minutes. The minutes will be to the left of the decimal point:

0.846153846 x 60
= 50.76923076
= 50 minutes

3) Find the number of seconds by multiplying what is remaining from step 2 by 60 seconds. The seconds will be to the left of the decimal point:

0.76923076 x 60
= 46.1538456
= 46 seconds

So 3 hours 50 mins and 46 seconds

3 0

3 years ago