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Sidana [21]
4 years ago
13

What is the distance from The sun to Jupiter?

Physics
1 answer:
Tomtit [17]4 years ago
8 0

The distance between the Sun and the Jupiter is nearly 779 million kilometer.

<u>Explanation: </u>

Jupiter is approximately 778.5 million kilo meter from the Sun, it is in approximation of 484 million miles. To be exact in separating distance between the two, it is 778547200 kilo meters.

The distance measured is an average taken due to the elliptical path undertaken by the planet for its planetary motion around the Sun, according to the Kepler’s law of planetary motion.

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What was the maximum speed of the car in your experiment?
MA_775_DIABLO [31]
The measurements used in the experiment is the amount of speed over time.

The measurement of speed is indicated along the “y” axis.

Upon viewing the graph, the highest point along the “y” axis shown is 25 m/s. This would be the maximum.

The maximum speed of the car would be 25 m/s.
7 0
3 years ago
State any five branches of physics​
prisoha [69]

Answer:

  1. Thermodynamics
  2. Quantum mechanics
  3. Nuclear physics
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3 years ago
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Why must the operating temperature of a heat engine be higher than that of the cold sink?
Anna35 [415]

Answer:

It is important to note, that the 2nd Law of thermodynamics plays no fundamental role in answering this question; we need a heat sink because the entropy is a state function, and at the end of the reversible process (which is visualized through the Carnot cycle diagram relevant for this problem), the entropy value of the system must return to the value it had originally.

6 0
4 years ago
The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The t
Gemiola [76]

Answer:

\omega=7.16*10^{-13}\frac{rad}{s}

Explanation:

The angular speed is given by:

\omega=\frac{v}{r}

Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to \frac{m}{s}:

1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}

Now, we calculate the angular speed:

\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}

8 0
4 years ago
If a running system has a total change in heat of 295 joules, and it's running at a temperature of 402 kelvin, what is the entro
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\[S = \Delta Q / T\] where S - entropy, Q heat and T temperature \[S = 295/402 = 0.733 \]
SO IT'S B 
8 0
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