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sammy [17]
3 years ago
12

What is the approximate number of galaxies in the known universe, assuming that the milky way has an average mass and that all o

f the mass in the known universe is in galaxies? the mass of the milky way is on the order of 1042 kg, and the mass of the known universe is about about 1053 kg?
Physics
1 answer:
igor_vitrenko [27]3 years ago
5 0
I think that the correct reporting of the given values are 10⁴² kg for the Milky Way and 10⁵³ kg for the whole universe. That would be logical in that case. Since it is mentioned that each galaxy has the same mass as the Milky Way, then the number of galaxies in the universe are:

Number of galaxies = 10⁵³ kg/10²⁴ kg = 10¹¹ or 100 trillion galaxies 
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The diagram shows a position-time graph What is the displacement of the object
algol13

The object is moving, so at different times, it has different displacement.  I'm guessing that you probably want to know the displacement at the end of the time on the graph ... 5 seconds.

Displacement is the distance and the direction FROM (the position at the  beginning) TO (the position at the end).

At the beginning ... time=0 ... the position is 1 meter.

At the end ... time=5 ... the position is zero.

The distance FROM the beginning TO the end is (zero - 1m) .  That's  <em>-1m </em>.


5 0
3 years ago
An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
oksian1 [2.3K]

Answer:

T=0.372 s, f=2.7 Hz, w=16.9 rad/s, k=179.2 N/m, v= 8.78 m/s, F= 48.4 N

Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

b)

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

d)

Spring Constant:

As w=\sqrt{k/m}

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

⇒ v= 8.78 m/s

f)

Maximum force will be exerted on the block when it is at maximum distance.

F= k* A   ( A is amplitude of oscillation)

F= 179.2 * 0.27 N

F= 48.4 N

5 0
3 years ago
Which of the following are results of the force of gravity?
docker41 [41]
Hello,

Here is your answer:

The proper answer for this question is option B "When released,a book falls to the ground". That's because of gravity the book will hit the ground!

Your answer is B.

If you need anymore help feel free to ask me!

Hope this helps.
7 0
3 years ago
Read 2 more answers
Why is potassium and sodium considered as reactive metals?​
boyakko [2]

Answer:

because they are found freely in nature uncombined so they are highly reactive with other elements

3 0
3 years ago
Read 2 more answers
Proyektil dengan massa 10 gram ditembak dari senapan dengan massa 5 kg. Kecepatan proyektil sesaat setelah memotret dari laras a
notka56 [123]

Answer:

-0.4 m/s

Explanation:

According to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.

The total momentum before the shot is zero, since they are both at  rest:

p_i = 0

While the total momentum after the shot can be written as:

p_f = mv+MV

where

m = 10 g = 0.010 kg is the mass of the bullet

M = 5 kg is the mass of the rifle

v = 200 m/s is the velocity of the bullet

V is the recoil velocity of the rifle

Since the total momentum is conserved, we can write:

p_i = p_f

So

0=mv+MV

And solving for V, we find the recoil velocity:

V=-\frac{mv}{M}=-\frac{(0.010)(200)}{5}=-0.4 m/s

and the negative sign indicates that the velocity is opposite to the bullet.

7 0
3 years ago
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