All categories

2 years ago

What is in the center of our galaxy?

Physics

2 answers:

irakobra [83]2 years ago

8 0

Answer:

The Galactic center also known as "A supermassive black hole"

Explanation:

"Its the rotational center of the Milky way" -google

sattari [20]2 years ago

7 0

The answer would be the sun

You might be interested in

A distant galaxy has a redshift z = 5.82 and a recessional velocity vr = 287,000 km/s (about 96% of the speed of light.) Notice

vlabodo [156]

Answer: 4100 Mpc

Explanation:

Since H o = 70 km/s/Mpc

Redshift z = 5.82

Recessional velocity vr = 287,000 km/s

Then, the distance to the galaxy in light years will be:

= Recessional velocity / H o

= 287000 / 70

= 4100 Mpc

7 0

3 years ago

Which two types of air masses would likely form a subtropical jet stream?

shepuryov [24]

The exact location of the 'subtropical-jet stream' is located at the North phase of 30 Degrees and the reason behind this is due to the variation of air which lies on the region of mid-altitude and warmer equatorial air. the correct answer would be Cool and Warm air masses meeting near the equator.

5 0

2 years ago

Read 2 more answers

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

8 0

2 years ago

If a boat and its riders have a mass of 900 kg and the boat drifts in at 1.4 m/s how much work does sam do to stop it?

Rama09 [41]

The initial kinetic energy of the boat and its rider is
$K_i = \frac{1}{2} mv_i^2 = \frac{1}{2}(900 kg)(1.4 m/s)^2=882 J$

After Sam stops it, the final kinetic energy of the boat+rider is
$K_f = 0 J$
because its final velocity is zero.

For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
$W=K_f-K_i =0-882 J=-882 J$
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.

6 0

2 years ago

Please can someone give a clear explanantion, <br><br> no extra links thanks

Tems11 [23]

Answer:

the extension recorded by the student would be smaller than the actual extension of the spring

3 0

2 years ago