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2 years ago

Differences between Light year and Cosmic year in two points

Physics

1 answer:

rusak2 [61]2 years ago

3 0

Answer:

A cosmic year is 365.25 days, some times called a side real year and is just the time it takes for us to go round the sun once.

A light year is the distance light travels in a year. Now light travels at about 186,262 miles a Second! Which is not slow by any ones book.

An experiment was conducted just after Christmas a few years ago. Two girls were selected from the audience and went into two phone boxes a few feet apart. They could only hear each other via the phones. The phone call went to a ground station about 200 miles away, then up to a geostationary coms satellite, back to a ground station 1/3 of the way around the world, then repeated, with a third satellite before being sent from another ground station back to London and the other phone box. We the audience could hear both sides of the conversation from both boxes. And could hear the delay between sending and receiving. So even at the speed of light, there was about 1.5 seconds of delay. So because distances in space are so vast that saying a star is x millions of miles away causes problems, you run out of zero’s! So our nearest other star is about 4.5 light years away. Our sun (our nearest start) is about 8 light minuets away. Varies slightly as our orbit is not 100% cirular.

I HOPE THIS IS HELPFUL.

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When light is directed on a metal surface, the kinetic energies of the photoelectrons a) are random b) vary with the frequency o

jekas [21]

Answer:

b) vary with the frequency of the light

Explanation:

The phone electric effect can be expressed as

K.E=(hv -W•)

Where K.E is the Kinectic energy

W• = work function of the metal

ν =frequency of the radiation

h = Planck's constat

Then, we can see that K.E is proportional linearly to "v" in the equation above.

Therefore, When light is directed on a metal surface, the kinetic energies of the photoelectrons vary with the frequency of the light

5 0

2 years ago

A sinusoidal transverse wave of amplitude ym = 8.4 cm and wavelength = 5.3 cm travels on a stretched cord. Find the ratio of the

Scilla [17]

Answer:

The ratio is 9.95

Solution:

As per the question:

Amplitude, $y_{m} = 8.4\ cm$

Wavelength, $\lambda = 5.3\ cm$

Now,

To calculate the ratio of the maximum particle speed to the speed of the wave:

For the maximum speed of the particle:

$v_{m} = y_{m}\times \omega$

where

$\omega = 2\pi f$ = angular speed of the particle

Thus

$v_{m} = 2\pi fy_{m}$

Now,

The wave speed is given by:

$v = f\lambda$

Now,

The ratio is given by:

$\frac{v_{m}}{v} = \frac{2\pi fy_{m}}{f\lambda}$

$\frac{v_{m}}{v} = \frac{2\pi \times 8.4}{5.3} = 9.95$

8 0

3 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$