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2 years ago

The milky way galaxy is the galaxy that is closest to us contains the most stars is farthest from us contains our solar system

1 answer:

6 0

The Milky Way galaxy is the one that the sun is a member of, and it contains
our solar system. We're in it, and you can't get much closer than that.

The Milky Way is known to be bigger than your average galaxy, but it's
probably not correct to say that it contains the 'most' stars of any galaxy.
The estimate for the Milky Way is only a few hundred billion stars.

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The mass of an atom is ________________________.

alexandr402 [8]

The mass of an atom comes from the protons and neutrons that is found in the nucleus. The number of protons is the atomic number of an element. To find the number of neutrons, subtract the atomic number from the mass of an atom. For example, sodium’s atomic number is 11. This will tell us that sodium has 11 protons in it. The atomic mass of sodium is 23. So subtract 23 form 11 gives us 12. Therefore, there are 12 neutrons in sodium.

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3 years ago

Surface currents and waves are powered by what

Allisa [31]

It is powered by the Earth's rotation and the moon gives a little boost.

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2 years ago

An unstretched spring has a length of 0.30 m. When the spring is stretched to a total length of 0.60 m, it supports traveling wa

IceJOKER [234]

Explanation:

Below is an attachment containing the solution

7 0

2 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

3 years ago

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of thebuilding. Ignore air resis

Bumek [7]

Answer:

a)3.5s

b)28.57m/S

c)34.33m/S

d)44.66m/S

Explanation:

Hello!

we will solve this exercise numeral by numeral

a) to find the time the ball takes in the air we must consider that vertically the ball experiences a movement with constant acceleration whose value is gravity (9.81m / S ^ 2), that the initial vertical velocity is zero, we use the following equation for a body that moves with constant acceleration

$Y= VoT+0.5gt^{2}$

where

Vo = Initial speed =0

T = time

g=gravity=9.81m/s^2

y = height=60m

solving for time

$Y=0.5gt^2\\t=\sqrt{\frac{Y}{0.5g} } \\t=\frac{60}{0.5(9.81)} \\$

T=3.5s

b)The horizontal speed remains constant since there is no horizontal acceleration. with the value of the distance traveled (100m) and the time that lasts in the air (3.5s) we estimate the horizontal speed

$V=\frac{x}{t} =\frac{100}{3.5}=28.57m/s$

to find the final vertical velocity we use the equations for motion with constant velocity as follows

Vf=Vo+g.t

Vf=0+(9.81 )(3.5)=34.335m/S

d)Finally, to find the resulting velocity, we add the horizontal and vertical velocities vectorially, this is achieved by finding the square root of the sum of its squares

$V=\sqrt{Vx^2+Vy^2} =\sqrt{34.33^2+28.57^2} =44.67m/S$

7 0

2 years ago