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3 years ago

Differences between Light year and Cosmic year in two points

Physics

1 answer:

rusak2 [61]3 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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The function x = (1.2 m) cos[(3πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 9.7 s, what are the (a) di

nlexa [21]

Answer and Explanation:

Let:

$x(t)=Acos(\omega t+ \phi)$

The equation representing a simple harmonic motion, where:

$x=Displacement\hspace{3}from\hspace{3}the\hspace{3}equilibrium\hspace{3}point\\A=Amplitude \hspace{3}of\hspace{3} motion\\\omega= Angular \hspace{3}frequency\\\phi=Initial\hspace{3} phase\\t=time$

As you may know the derivative of the position is the velocity and the derivative of the velocity is the acceleration. So we can get the velocity and the acceleration by deriving the position:

$v(t)=\frac{dx(t)}{dt} =- \omega A sin(\omega t + \phi)\\\\a(t)=\frac{dv(t)}{dt} =- \omega^2 A cos(\omega t + \phi)$

Also, you may know these fundamental formulas:

$f=\frac{\omega}{2 \pi} \\\\T=\frac{2 \pi}{\omega}$

Now, using the previous information and the data provided by the problem, let's solve the questions:

(a)

$x(9.7)=1.2 cos((3 \pi *(9.7))+\frac{\pi}{5} ) \approx -0.70534m$

(b)

$v(9.7)=-(3\pi) (1.2) sin((3\pi *(9.7))+\frac{\pi}{5} ) \approx 9.1498 m/s$

(c)

$a(9.7)=-(3 \pi)^2(1.2)cos((3\pi*(9.7))+\frac{\pi}{5} )\approx -62.653m/s^2$

(d)

We can extract the phase of the motion, the angular frequency and the amplitude from the equation provided by the problem:

$\phi = \frac{\pi}{5}$

(e)

$f=\frac{\omega}{2 \pi} =\frac{3\pi}{2 \pi} =\frac{3}{2} =1.5 Hz$

(f)

$T=\frac{2 \pi}{\omega} =\frac{2 \pi}{3 \pi} =\frac{2}{3} \approx 0.667s$

8 0

3 years ago

Read 2 more answers

A 1.2 x10 3 kilogram automobile in motion strikes a 1.0 x 10 -4 kilogram insect as a result the insect is accelerated at a rate

Mariana [72]

According to Newton's second law, the force applied to an object is equal to the product between the mass of the object and its acceleration:
$F=ma$
where F is the magnitude of the force, m is the mass of the object and a its acceleration.

In this problem, the object is the insect, with mass $m=1.0 \cdot 10^{-4} kg$ . The acceleration of the insect is $a=1.0 \cdot 10^2 m/s^2$ , therefore we can calculate the force exerted by the car on the insect:
$F=ma=(1.0 \cdot 10^{-4} kg)(1.0 \cdot 10^2 m/s^2)=0.01 N$

How do we find the force exerted by the insect on the car?
According to Newton's third law (known as action-reaction law), when an object A exerts a force on an object B, object B also exerts a force equal and opposite on object A. Therefore, the force exerted by the insect on the car is equal to the force exerted by the car on the object, so it is 0.01 N.

6 0

3 years ago

It took 1700 J of work to stretch a spring from its natural length of 2m to a length of 5m. Find the spring's force constant.

Alenkasestr [34]

The work to stretch a spring from its rest position is

   (1/2) (spring constant) (distance of the stretch)²

   E = 1/2 k x² .

You said it takes 1700 joules to stretch the spring 3 meters from its rest position, so we can write

   1700 joules = 1/2 k (3m)²

1 joule = 1 newton-meter

   1700 N-m = 1/2 k (3m)²

Multiply each side by 2: 3400 N-m = k · 9m²

Divide each side by 9m²    k = 3400 N-m / 9m²

   = (377 and 7/9)   newton per meter

7 0

3 years ago

Read 2 more answers

A charge feels a 7.89 x 10-7 N force when it moves 2090 m/s at a 29.4° angle to a 4.23 x 10-3 T magnetic field. What is the amou

tamaranim1 [39]

We have that the amount of the charge q is

$q=1.8*10^{-7}$