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Luba_88 [7]
3 years ago
9

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

object; the ring has a diameter of about 18 light-years and an orbital speed of about 180 km/s. answers
Physics
1 answer:
Setler [38]3 years ago
5 0

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

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Which wave has a longer period and how many times, the wave with a frequency of 7000Hz or the wave with a frequency of 21.000Hz?
Nuetrik [128]

Answer:

I would say the answer is the wave of 21.000Hz

Explanation:

Because it has more frequency, and as more frequency you add, the time or longer period also increases.

3 0
2 years ago
Enter an expression, in terms of defined quantities and g, for the force that the scale under left pillar shows
ella [17]

The expression, in terms of defined quantities and g is therefore Fu =((mg/2) +2 mp) g

<h3>What is a Scale?</h3>

This can be defined as a balance or any of various other instruments or devices for weighing.

The expression in terms of defined quantities and g, for the force that the scale under left pillar shows that Fu =((mg/2) +2 mp) g .

Read more about Force here brainly.com/question/4515354

5 0
2 years ago
A block weighing 400 kg rests on a horizontal surface and supports on top of it ,another block of weight 100 kg which is attache
Paladinen [302]

Answer:

F_a=1470\ N

Explanation:

<u>Friction Force</u>

When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.

There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.  

Please find the free body diagrams in the figure provided below.

The equilibrium condition for the mass 1 is

\displaystyle F_a-F_{r1}-F_{r2}=m.a=0

The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa

\displaystyle F_a=F_{r1}+F_{r2}.....[1]

The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is

\displaystyle F_{r2}-T=0

The friction forces are computed by

\displaystyle F_{r2}=\mu \ N_2=\mu\ m_2\ g

\displaystyle F_{r1}=\mu \ N_1=\mu(m_1+m_2)g

Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.

Replacing in [1]

\displaystyle F_{a}=\mu \ m_2\ g\ +\mu(m_1+m_2)g

Simplifying

\displaystyle F_{a}=\mu \ g(m_1+2\ m_2)

Plugging in the values

\displaystyle F_{a}=0.25(9.8)[400+2(100)]

\boxed{F_a=1470\ N}

8 0
3 years ago
A ball is tossed vertically upward. When it reaches its highest point (before falling back downward) Group of answer choices the
nignag [31]

Answer:

the velocity is zero, the acceleration is directed downward, and the force of gravity acting on the ball is directed downward

Explanation:

Is this exercise in kinematics

          v = v₀ - g t

where g is the acceleration of the ball, which is created by the attraction of the ball to the Earth.

At the highest point

velocity must be zero.

The acceleration depends on the Earth therefore it is constant at this point and with a downward direction.

The force of the earth on the ball is towards the center of the Earth, that is, down

all other alternatives are wrong

7 0
3 years ago
A student dips a strip of metal into a liquid. Which is evidence that only a physical change has occurred?
sattari [20]

Answer:

PLS add the picture too.....

3 0
3 years ago
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