What source of energy is responsible for the water cycle

Astronomers have observed a small, massive object at the center of our Milky Way Galaxy. A ring of material orbits this massive

Setler [38]

Answer:

The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

Explanation:

Given that,

Diameter = 10 light year

Orbital speed = 180 km/s

Suppose determine the mass of the massive object at the center of the Milky Way galaxy.

Take the distance of one light year to be 9.461×10¹⁵ m. I was able to get this it is 4.26×10³⁷ kg.

We need to calculate the radius of the orbit

Using formula of radius

$r=\dfrac{d}{2}$

$r=\dfrac{15\times9.461\times10^{15}}{2}$

$r=7.09\times10^{16}\ m$

We need to calculate the mass of the massive object at the center of the Milky Way galaxy

Using formula of mass

$M=\dfrac{v^2r}{G}$

Put the value into the formula

$M=\dfrac{(180\times10^3)^2\times7.09\times10^{16}}{6.67\times10^{-11}}$

$M=3.44\times10^{37}\ Kg$

Hence, The mass of the massive object at the center of the Milky Way galaxy is $3.44\times10^{37}\ Kg$

5 0

2 years ago

Ocean waves of wavelength 26 m are moving directly toward a concrete barrier wall at 4.0 m/s . The waves reflect from the wall,

Genrish500 [490]

Answer:

a) the distance between her and the wall is 13 m

b) the period of her up-and-down motion is 6.5 s

Explanation:

Given the data in the question;

wavelength λ = 26 m

velocity v = 4.0 m/s

a) How far from the wall is she?

Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;

x = λ/2

we substitute

x = 26 m / 2

x = 13 m

Therefore, the distance between her and the wall is 13 m

b) What is the period of her up-and-down motion?

we know that the relationship between frequency, wavelength and wave speed is;

v = fλ

hence, f = v/λ

we also know that frequency is expressed as the reciprocal of the time period;

f = 1/T

Hence

1/T = v/λ

solve for T

Tv = λ

T = λ/v

we substitute

T = 26 m / 4 m/s

T = 6.5 s

Therefore, the period of her up-and-down motion is 6.5 s

6 0

2 years ago

An airplane starts from rest and accelerates at a constant rate of 3.00 m/s2 for 30.0 s before leaving the ground. a. How far di

rosijanka [135]

The formula we can use in this case is:

d = v0t + 0.5 at^2

v = at + v0

where,

d = distance travelled

v0 = initial velocity = 0 since at rest

t = time travelled

a = acceleration

v = final velocity when it took off

a. d = 0 + 0.5 * 3 * 30^2

d = 1350 m

b. v = 3 * 30 + 0

8 0

3 years ago

Add vectors 7.5 m/s and the vector -2.9 m/s

densk [106]

How To:

How do you add two vectors?

To add or subtract two vectors, add or subtract the corresponding components. Let →u=⟨u1,u2⟩ and →v=⟨v1,v2⟩ be two vectors. The sum of two or more vectors is called the resultant. The resultant of two vectors can be found using either the parallelogram method or the triangle method .

5 0

2 years ago

What is the focal length of the eye-lens system when viewing an object at infinity? assume that the lens-retina distance is 2.0

34kurt

The focal length of the eye-lens system will be exactly 2.0 cm. The lens thickness is altered depending on the distance of the object being viewed so that the focal length is ALWAYS equal to the lens-retina distance. If this is not true fr an individual, they need glasses :)

4 0

3 years ago