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2 years ago

How many stars are in our galaxy??

Physics

2 answers:

Firdavs [7]2 years ago

8 0

Answer:

Explanation:

There are 400 billion stars in our galaxy

Vesnalui [34]2 years ago

3 0

Answer:

Hi! I believe this is your answer:

400 billion

Hope this helps, sorry if it's wrong!!

Explanation:

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Describe what it means to view something from a frame of reference. Give an example to illustrate your explanation.

Komok [63]

Answer:

In physics, the concept of a frame of reference is used to specify the perspective from which an object or event is observed.

A frame of reference is where the measurements or observations will be made. Because of this, observing an event may be different when changing from one frame of reference to another, because the measurements will be different.

Defining frames of reference is necessary because the movement is relative, it may be that from our perspective or from a frame of reference on earth we are at rest, but seen from a frame of reference in space, we are in motion due that the earth is always moving.

Another example of frames of reference is a moving plane. Seen from the ground an object in the plane moves at the speed at which the plane travels, but if the frame of reference is fixed on the plane, the object is at rest.

3 0

3 years ago

A laboratory electromagnet produces a magnetic field of magnitude 1.50T . A proton moves through this field with a speed of 6.00

hram777 [196]

The magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

The magnitude of the force on a charged particle moving in a magnetic field is given by the formula,

F= qvB

Here q is the charge on proton = 1.6 x 10^-19 C.

v is the velocity with which the particle is moving = 6.00 x 10^6 m/s

And B is the value of the magnetic field = 1.5 T

Putting the given values in the above equation,

F = 1.6 x 10^-19 x 6 x 10^6 x 1.5 = 1.44 x 10^-12 N.

Hence, the magnitude of maximum magnetic force that could be exerted on the proton is 1.44 x 10^-12 N.

To know more about "magnetic force", refer to the link given below:

brainly.com/question/13791875?referrer=searchResults

#SPJ4

3 0

1 year ago

As a skateboarder moves downhill, some of the energy of the skateboarder is

Sonja [21]

As the skateboard rolls down the ramp it loses potential energy and gains kinetic energy.

6 0

3 years ago

Read 2 more answers

Is a force that opposes the motion between two objects in contact with each other.

Sergeeva-Olga [200]

Answer: Friction :)

7 0

2 years ago

Read 2 more answers

Compare the circular velocity of a particle orbiting in the Encke Division, whose distance from Saturn 133,370 km, to a particle

Ket [755]

Answer:

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

Explanation:

The circular velocity is define as:

$v = \frac{2 \pi r}{T}$

Where r is the radius of the trajectory and T is the orbital period

To determine the circular velocity of both particles it is necessary to know the orbital period of each one. That can be done by means of the Kepler’s third law:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

Case for the particle in the Encke Division:

$T^{2} = r^{3}$

$T = \sqrt{(133370 Km)^{3}}$

$T = \sqrt{(2.372x10^{15} Km)}$

$T = 4.870x10^{7} Km$

It is necessary to pass from kilometers to astronomical unit (AU), where 1 AU is equivalent to 150.000.000 Km ( $1.50x10^{8} Km$ )

1 AU is defined as the distance between the earth and the sun.

$\frac{4.870x10^{7} Km}{1.50x10^{8}Km} . 1AU$

$T = 0.324 AU$

But 1 year is equivalent to 1 AU according with Kepler’s third law, since 1 year is the orbital period of the earth.

$T = \frac{0.324 AU}{1 AU} . 1 year$

$T = 0.324 year$

That can be expressed in units of days

$T = \frac{0.324 year}{1 year} . 365.25 days$

$T = 118.60 days$

Circular velocity for the particle in the Encke Division:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (133370 Km)}{(118.60 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$118.60 days .\frac{86400 s}{1 day}$ ⇒ 10247040 s

$133370 Km .\frac{1000 m}{1 Km}$ ⇒ 133370000 m

$v = \frac{2 \pi (133370000 m)}{(10247040 s)}$

$v = 81.778 m/s$

Case for the particle in the D Ring:

For the case of the particle in the D Ring, the same approach used above can be followed

$T^{2} = r^{3}$

$T = \sqrt{(69000 Km)^{3}}$

$T = \sqrt{(3.285x10^{14} Km)}$

$T = 1.812x10^{7} Km$

$\frac{1.812x10^{7} Km}{1.50x10^{8}Km} . 1 AU$

$T = 0.120 AU$

$T = \frac{0.120 AU}{1 AU} . 1 year$

$T = 0.120 year$

$T = \frac{0.120 year}{1 year} . 365.25 days$

$T = 43.83 days$

Circular velocity for the particle in D Ring:

$v = \frac{2 \pi r}{T}$

$v = \frac{2 \pi (69000 Km)}{(43.83 days)}$

For a better representation of the velocity, kilometers and days are changed to meters and seconds respectively.

$43.83 days . \frac{86400 s}{1 day}$ ⇒ 3786912 s

$69000 Km . \frac{1000 m}{ 1 Km}$ ⇒ 69000000 m

$v = \frac{2 \pi (69000000 m)}{(3786912 s)}$

$v = 114.483 m/s$

$\frac{114.483 m/s}{81.778 m/s} = 1.399$

The particle in the D ring is 1399 times faster than the particle in the Encke Division.

7 0

2 years ago