In order to make his measurements for determining the Earth-Sun distance, Aristarchus waited for the Moon's phase to be exactly half full while the Sun was still visible in the sky. For this reason, he chose the time of a half (quarter) moon.
<h3 /><h3>How did Aristarchus calculate the distance to the Sun?</h3>
It was now possible for another Greek astronomer, Aristarchus, to attempt to determine the Earth's distance from the Sun after learning the distance to the Moon. Aristarchus discovered that the Moon, the Earth, and the Sun formed a right triangle when they were all equally illuminated. Now that he was aware of the distance between the Earth and the Moon, all he needed to know to calculate the Sun's distance was the current angle between the Moon and the Sun. It was a wonderful argument that was weakened by scant evidence. Aristarchus calculated this angle to be 87 degrees using only his eyes, which was not far off from the actual number of 89.83 degrees. But when there are significant distances involved, even slight inaccuracies might suddenly become significant. His outcome was more than a thousand times off.
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Answer:
b) total energy input equals total energy output
Explanation:
The first law of thermodynamics is a generalization of the conservation of energy in thermal processes. It is based on Joule's conclusion that heat and energy are equivalent. But to get there you have to get around some traps along the way.
From Joule's conclusion we might be tempted to call heat "internal" energy associated with temperature. We could then add heat to the potential and kinetic energies of a system, and call this sum the total energy, which is what it would conserve. In fact, this solution works well for a wide variety of phenomena, including Joule's experiments. Problems arise with the idea of heat "content" of a system. For example, when a solid is heated to its melting point, an additional "heat input" causes the melting but without increasing the temperature. With this simple experiment we see that simply considering the thermal energy measured only by a temperature increase as part of the total energy of a system will not give a complete general law.
Instead of "heat," we can use the concept of internal energy, that is, an energy in the system that can take forms not directly related to temperature. We can then use the word "heat" to refer only to a transfer of energy between a system and its environment. Similarly, the term work will not be used to describe something contained in the system, but describes a transfer of energy from one system to another. Heat and work are, therefore, two ways in which energy is transferred, not energies.
In an isolated system, that is, a system that does not exchange matter or energy with its surroundings, the total energy must remain constant. If the system exchanges energy with its environment but not matter (what is called a closed system), it can do so only in two ways: a transfer of energy either in the form of work done on or by the system, either in the form of heat to or from the system. In the event that there is energy transfer, the change in the energy of the system must be equal to the net energy gained or lost by the environment.
Answer:
Answer for the question is given in the attachment.
Explanation:
Answer:
0.78 m
Explanation:
By the conservation of energy, the energy that they gain from potential energy, must be equal to the kinetic energy. So, for Adolf:
Ep = Ek
ma*g*ha = ma*va²/2
Where ma is the mass of Adolf, g is the gravity acceleration (10 m/s²), ha is the height that he reached, and va is the velocity. So:
100*10*0.51 = 100*va²/2
50va² = 510
va² = 10.2
va = √10.2
va = 3.20 m/s
Before the push, both of them are in rest, so the momentum must be 0. The system is conservative, so the momentum after the push must be equal to the momentum before the push:
ma*va + me*ve = 0, where me and ve are the mass and velocity of Ed. So:
100*3.20 + 81ve = 0
81ve = 320
ve = 3.95 m/s
By the conservation of energy for Ed:
me*g*he = me*ve²/2
81*10*he = 81*(3.95)²/2
810he = 631.90
he = 0.78 m
