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ale4655 [162]
3 years ago
12

A star rotates in a circular orbit about the center of its galaxy. The radius of the orbit is 6.9 x 1020 m, and the angular spee

d of the star is 9.2 x 10-15 rad/s. How long (in years) does it take for the star to make one revolution around the center?
Physics
1 answer:
andrew-mc [135]3 years ago
8 0

Answer:

2.16 x 10^7 years

Explanation:

R = 6.9 x 10^20 m

ω = 9.2 x 10^-15 rad/s

T = 2π / ω

T = ( 2 x 3.14 ) / ( 9.2 x 10^-15)

T = 6.826 x 10^14 second

T = 2.16 x 10^7 years

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rodikova [14]
<span>Because of the orbit of the earth and the sun and the moon. </span>
6 0
3 years ago
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In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache
Masteriza [31]

Answer:

b) True.    the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

     fr - w = ma

Where the friction force has some form of type.

     fr = G v + H v²

Let's replace

     (G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

      fr - w = 0

      fr = mg

      (G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

     G v = mg

     v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0
3 years ago
A. 24.89<br> B. 25.89<br> C. 17.74<br> D. 19.73
Veronika [31]

Answer: D

Explanation:

Just did it got an 100

5 0
2 years ago
you place a beaker of water on a hot plate and heat it up eventually it starts to boil and you see bubbles for me you also seeme
Nuetrik [128]
It's a physical change. the water is changing it's physical state

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3 years ago
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