There is no air in space astronauts in space cannot hear sounds from outside their spacesuits explain this

Allisa [31]

Answer:

positions of motion is the answer as moving is a motion

3 0

1 year ago

Suppose your local apparent solar time is 11 am, and you have a clock that says it is noon in Greenwich. What is your longitude

jeyben [28]

The longitude based on the time difference is 15 degrees.

<h3>Longitude of complete rotation of the Earth</h3>

The longitude of a complete rotation of the earth in a 24 hours is calculated as follows;

$= \frac{360^0}{24} = 15^0 \ of \ longitude$

<h3>Time difference</h3>

The time difference between the local apparent solar time and the Greenwich time is calculated as follows;

$t = 12 pm \ - \ 11 am\\\\t = 1 \ hour$

Since it is one hour time difference, the longitude is 15 degrees.

Learn more about Earth longitude here: brainly.com/question/1939015

5 0

2 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

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

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

2 years ago

A stone is thrown horizontally from the top of an inclined plane (angle of inclination θ). How would I find the initial speed of

Katen [24]

Answer:

S = V t where S is the horizontal distance traveled

1/2 g t^2 = H where H is the vertical distance traveled

t^2 = 2 H / g

V^2 = S^2 / t^2 = S^2 g / (2 H) combining equations

tan theta = H / S

V^2 = S g / (2 tan theta)

Using S = L cos theta

V^2 = L g cos theta / (2 tan theta)

Giving V in terms of L and theta

7 0

2 years ago

here is a true or false Question pl,ease help..... . In a closed system, the volume of a gas is not related to the volume of its

KIM [24]

There are four states of matter: solid, liquid, gas and plasma. Solid has a shape and volume of its own. A liquid takes the shape of the container does not occupy entirely the volume of the container. For gas, it takes the shape of the container as well as taking the volume of the container. In this case, if the system is only composed of gas, the statement given is true.

5 0

2 years ago

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