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Volgvan
3 years ago
12

Astronomers discover a bright X-ray source in the Milky Way. They see no counterpart in the optical bands, but it has another st

ar orbiting it, and they determine that the X-ray source has a mass of 10 MSun. What is the most likely origin of the X-rays?
a. a black hole
b. a neutron star
c. a white dwarf
d. a gamma-ray burst
Physics
1 answer:
VladimirAG [237]3 years ago
8 0

Answer:

a. a black hole

Explanation:

X-ray emission from the central degrees of the Milky Way Bright X-ray emission traces the coherent edge brightened shell-like feature, dubbed the northern chimney, located north of Sgr A* and characterized by a diameter of about 160 pc. On the opposite side, the southern chimney appears as a bright linear feature. Bright X-ray emission is observed at high latitude

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Which of the following statements is true?
Bad White [126]
The answer is c hope this helps



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Part B For this activity, you'll analyze the motion of the marked spot on the skateboarder's leg. For each dropdown labeled “cm”
Anika [276]

Answer:

Explanation:

c. By using the Select Data button and the Select Data Source optionExplanation:A scatter plot is a plot which is used to plot the points of the data on the horizontal and the vertical axis also it depicts how one variable is affected by the another. After preparing the scatter plot to enter the data in the scatter plot we need to use the data button and then data source option so that the data could be entered in the scatter plothence, option c is correct

7 0
2 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
2 years ago
Discuss the limitations of using the Doppler shift to determine an object's speed.
pantera1 [17]

Answer and Explanation:

Limitation of Doppler shift :

The Doppler impact is relevant when the speeds of the wellspring of sound and spectator are considerably less than the speed of sound. The movement of both the spectator and the source is along a similar straight line.When movement is not in straight line or velocity is not much less than speed of light then we can not use Doppler shift

This is the limitation of Doppler shift to determine the object distance

3 0
3 years ago
Suppose our apparatus will only allow us to distinguish the first order bright fringe from the central bright spot if they are s
ozzi

Answer:

λ = 8.716 mm

Explanation:

Given:

- d = 10 cm

- Q >= 5 degrees

Find:

- Find the shortest wavelength of light for which this apparatus is useful

Solution:

- The formula that relates the split difference and angle of separation between successive fringes is given by:

                                            d*sin(Q) = n*λ

Where,

λ: wavelength

d: split separation

Q: angle of separation between successive fringes

m: order number.

- Since this apparatus only shows the first order light so m =1

- the shortest possible wavelength corresponds to:

                                            d*sin(Q) = λ

                                            λ = 0.1*sin(5)

                                            λ = 8.716 mm

3 0
3 years ago
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