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Komok [63]
3 years ago
12

Imagine an advanced civilization living on a planet orbiting at a distance of 10 AU (1.5 billion kilometers) from a close binary

star system that consists of a 15 MSun red giant star and a 10 MSun black hole. The black hole is surrounded by an accretion disk. Sometime within the next million years or so, the civilization's planet is likely to be doomed because ___
Physics
1 answer:
aksik [14]3 years ago
6 0

Answer:

Civilization's planet is most likely to be doomed because within the next million years the red giant star will possibly supernova.

Explanation:

Civilization's planet is most likely to be doomed because within the next million years the red giant star will possibly supernova. The supernova is the bursting of the star that has reached its end life. This leads to release of massive amount of energy, which can be more than the heat and energy released by the sun.

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Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o
Colt1911 [192]

Answer:

There is a loss of fluid in the  container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

\Delta V = \beta V_0 \Delta T

Where,

\Delta V =Change in volume

V_0 =Initial Volume

\Delta T = Change in temperature

\beta = coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

\Delta V_T = \Delta V_l - \Delta V_c

Where,

\Delta V_l= Change in the volume of liquid

\Delta V_c= Change in the volume of copper

Then replacing with the previous equation we have:

\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T

\Delta V = (\beta_l-\beta_c)V_0\Delta T

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

\beta_c = 51*10^{-6}/\°C

\beta_l = 400*10^{-6}/\°C

V_0 = 16L

\Delta T = (95\°C-10\°C)

Replacing we have that,

\Delta V = (\beta_l-\beta_c)V_0\Delta T

\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)

\Delta V = 0.475L

Therefore there is a loss of fluid in the container of 0.475L

6 0
3 years ago
Which best describes the motion of air particles when a transverse wave passes through them?
Minchanka [31]
C.
The particles move perpendicular to the direction of the wave.
3 0
2 years ago
Star a has a parallax angle of 0.02 arcseconds, and star b has a parallax of angle of 0.1 arcseconds. which star is more distant
Studentka2010 [4]
<span>Star a is more distant and is approximately 5 times as far away as star b Parallax is the change in angle that one must do in order to observe the same object from different locations. The further away an object is, the smaller the parallax is. As the angles approach zero, the trig functions tend to be fairly linear. And 0.1 arc seconds and 0.02 arc seconds are close enough to zero for this to hold true. Since the parallax for star a is smaller than the parallax for star b, it is the more distant star. And since 0.1 divided by 0.02 = 5, it is approximately 5 times further away than star b.</span>
3 0
3 years ago
(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th
julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

V = -\dfrac{kQ}{r}

where k = 9\times 10^9 \text{ F/m}

Difference in potential between the points is

kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}

PD = 135\times 10^3\text{ V} = 135\text{ kV}

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]

10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10

\dfrac{1}{x} = 0

x = \infty

The charge chould be moved to infinity

7 0
3 years ago
A suspended object A is attracted to a charged object B, can one conclude that A is charged? Explain
irinina [24]

Explanation:

In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B

Example =

If Object B is Negatively Charged, the Object A should be Positively Charged

If the Object B is Positively Charged, the Object A should be Negatively Charged

Sometimes it can Mix as a Neutral as well

Hope this Helps

5 0
2 years ago
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