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3 years ago

What evidence supports the theory that there is a black hole at the center of our galaxy?

Physics

1 answer:

777dan777 [17]3 years ago

3 0

Explanation:

Scientist and astronomers have observed that large gas clouds and massive stars are orbiting around in an accelerated manner in center of our galaxy that is milky way. A massive star S2's motion has been studied and it is found that the star is revolving around the center of the galaxy. From this scientist have confirmed presence of super massive black hole of 3 million solar masses lurking around the center of milky way.

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A car travels at an average speed of 60km/h for 15 minutes. How far does the car travel in this time?

jok3333 [9.3K]

Answer:

In 15 minutes the car travels a distance of 15 km.

Explanation:

4 0

2 years ago

Read 2 more answers

Which of the following has the most momentum?

777dan777 [17]

Answer:

0.50 kg moving at 10 m/s

Explanation:

mv = p

0.50(10) = 5.0 kg•m/s

5.0(0.50) = 2.5 kg•m/s

1(2) = 2 kg•m/s

0.40(10) = 4.0 kg•m/s

5 0

2 years ago

Which of the following is true about the diameter of a circle?

Natalka [10]

Answer:

<em>option</em><em> A</em>

Explanation:

the diameter is twice the length of the radius

hope it helps

6 0

3 years ago

An astronaut is being tested in a centrifuge. The centrifuge has a radius of 11.0 m and, in starting, rotates according to θ = 0

Mazyrski [523]

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

$\theta=0.260t^2$

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

$\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t$

Then, you evaluate for t=2.40 s:

$\omega=0.52(2.40)=1.248\frac{rad}{s}$

(b) The linear velocity is calculated by using the following formula:

$v=\omega r$

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

$v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}$

(c) The tangential acceleration is:

$a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}$

(d) The radial acceleration is:

$a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}$

4 0

3 years ago

A cyclist, while overtaking another bike, increases his speed uniformly from 4.2 mis to 6.3 m/s over a time interval of 5.3 s. H

avanturin [10]

Answer:

27.82m

Explanation:

The distance of the cyclist can be calculated using the formula:

S = ut + 1/2 (at^2)

Where; S = distance

u = initial velocity

a = acceleration

t = time

However, to find acceleration, we use:

a = v - u/t, where v is the final velocity

According to this question, u = 4.2m/s, v = 6.3m/s, t = 5.3s

a = 6.3 - 4.2 / 5.3

a = 2.1/5.3

a = 0.396m/s^2

Hence, using S = ut + 1/2 (at^2)

S = (4.2 × 5.3) + 1/2 (0.396 × 5.3 × 5.3)

S = 22.26 + 5.56

S = 27.82

Therefore, the cyclist travel in 27.82m

8 0

2 years ago