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user100 [1]
3 years ago
11

Infrared observations of the orbits of stars close to the galactic center indicate a small object at the center with a mass of a

bout how many solar masses?
Physics
1 answer:
MariettaO [177]3 years ago
5 0

Answer:

(4.31±0.38) million Solar masses.

Explanation:

The galactic center is the center of the milky way around which the galaxy rotates. It is most likely the location of a supermassive black hole which has a mass of (4.31±0.38) million Solar masses. The location is called Sagittarius A*.

As there is interstellar dust in our line of sight from the Earth infrared observations need to be taken.

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A friend asks you how much pressure is in your car tires. You know that the tire manufacturer recommends 30 psi, but it's been a
Bezzdna [24]

Answer:

25 psi

Explanation:

The weight of the car is:

W = mg

W = 1550 kg * 9.8 m/s²

W = 15,190 N

Divided by 4 tires, each tire supports:

F = W/4

F = 15,190 N / 4

F = 3797.5 N

Pressure is force divided by area, so:

P = F / A

P = (3797.5 N) / (0.16 m × 0.14 m)

P ≈ 170,000 Pa

101,325 Pa is the same as 14.7 psi, so:

P ≈ 170,000 Pa × (14.7 psi / 101,325 Pa)

P ≈ 25 psi

8 0
3 years ago
Which is an example of a mixture?
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3 years ago
How to do this problem
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The answer I'm pretty sure is 3
4 0
3 years ago
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An 8.00 g bullet is fired into a 250 g block that is initially at rest at the edge of a table of 1.00 m height. The bullet remai
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5 0
3 years ago
A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc
max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }

f = \frac{1}{2\pi } ( 13.522)

f = 2.1535 Hz

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

#SPJ4

6 0
1 year ago
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