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Vsevolod [243]
3 years ago
6

Which law of thermodynamics would be violated if heat were to spontaneously flow between two objects of equal temperature?

Physics
1 answer:
maks197457 [2]3 years ago
5 0
<span>Actually the second law of thermodynamics would truly gets violated ie, which means that the entrophy changes of the isolated system can never be negative, which covers the above that if heat were to spontaneously flow between any two objects of equal temperature would be fully violated.</span>
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crimeas [40]
The movie? Or the book
4 0
3 years ago
EARTH SCIENCE PLEASE ANSWER
Rzqust [24]

The Earth Science answers are shown below.

Explanation:

1. The movement of the sun will change the angle it has on the sky in 30 minutes, it is always moving from the east to the west, so in 30 minutes it would move more west, no matter at what time you make the experiment. From Earth, the Sun looks like it moves across the sky in the daytime and appears to disappear at night. This is because the Earth is spinning towards the east. The Earth spins about its axis, an imaginary line that runs through the middle of the Earth between the North and South poles

2. No, both marks are the same distance from the ground.  the amount of stick above the mark will not affect the distance that the shadow of the mark moves at all. The Sun's clockwise motion is an apparent motion caused by the rotation of the Earth. The counterclockwise rotation of the Earth in the Sun's light causes the shadow of the gnomon to move clockwise. As the Sun appears to move higher above the horizon before solar noon, the shadow grows shorter and shorter.

3. In the summer the shadows are shorter, and in the winter the shadows are longer. In the morning your shadow will point west and in the afternoon it will point east. If your shadow is long, it is near sunrise or sunset. Your shadow is shortest around noon.

4. If the sun rises in the east and sets in the west, then the Earth should rotate in the opposite direction from west to east (anti-clockwise). Earth's spin (or rotation) on its axis. Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

4 0
4 years ago
The position-time graph of an object is found to be a straight line passing through the origin. What information about the motio
Nostrana [21]
-- The object either left or crossed the starting line exactly at time=0 .

-- The object has been traveling at constant speed for all time that
    we know about.
5 0
3 years ago
Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 403 km above the earth’s sur
BARSIC [14]

Answer:

v_A=7667m/s\\\\v_B=7487m/s

Explanation:

The gravitational force exerted on the satellites is given by the Newton's Law of Universal Gravitation:

F_g=\frac{GMm}{R^{2} }

Where M is the mass of the earth, m is the mass of a satellite, R the radius of its orbit and G is the gravitational constant.

Also, we know that the centripetal force of an object describing a circular motion is given by:

F_c=m\frac{v^{2}}{R}

Where m is the mass of the object, v is its speed and R is its distance to the center of the circle.

Then, since the gravitational force is the centripetal force in this case, we can equalize the two expressions and solve for v:

\frac{GMm}{R^2}=m\frac{v^2}{R}\\ \\\implies v=\sqrt{\frac{GM}{R}}

Finally, we plug in the values for G (6.67*10^-11Nm^2/kg^2), M (5.97*10^24kg) and R for each satellite. Take in account that R is the radius of the orbit, not the distance to the planet's surface. So R_A=6774km=6.774*10^6m and R_B=7103km=7.103*10^6m (Since R_{earth}=6371km). Then, we get:

v_A=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{6.774*10^6m} }=7667m/s\\\\v_B=\sqrt{\frac{(6.67*10^{-11}Nm^2/kg^2)(5.97*10^{24}kg)}{7.103*10^6m} }=7487m/s

In words, the orbital speed for satellite A is 7667m/s (a) and for satellite B is 7487m/s (b).

7 0
3 years ago
A tall, cylindrical chimney falls over when its base is ruptured. Treat the chimney as a thin rod of length 53.2 m. At the insta
rusak2 [61]
From an energy balance, we can use this formula to solve for the angular speed of the chimney

ω^2 = 3g / h sin θ

Substituting the given values:
ω^2 = 3 (9.81) / 53.2 sin 34.1
ω^2 = 0.987 /s

The formula for radial acceleration is:
a = rω^2

So,
a = 53.2 (0.987) = 52.494 /s^2

The linear velocity is:
v^2 = ar
v^2 = 52.949 (53.2) = 2816.887

The tangential acceleration is:
a = r v^2
a = 53.2 (2816.887)
a = 149858.378 m/s^2

If the tangential acceleration is equal to g:
g = r^2 3g / sin θ
Solving for θ
θ = 67° 
7 0
3 years ago
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