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3 years ago

What is the most widely believed though likely false aspect of the olympic truce story

Physics

2 answers:

artcher [175]3 years ago

3 0

Answer:

The truce ended all wars

Explanation:

The Olympic Truce or the sacred truce, is the declaration made by ambassadors for the laying down of arms and the stop of all hostilities starting from seven days before the Olympic games began and ending seven days after the games to provide safe passage for travelling athletes and also spectators to and from the Olympic games

However, the games were held during Persian wars and also during the Peloponnesian wars. As such the truce stopped mainly conflicts that have direct impact on the Olympic games and there was an attack in Olympia in 364 BC during the games.

ehidna [41]3 years ago

3 0

Answer:

The truce ended all wars

Explanation:

Apex

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Answer:

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2 years ago

Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa

Alex

Answer:

$B=1.1636*10^{-3}T$

Explanation:

Given data

$d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\$

To find

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3 years ago

An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/

Katen [24]

<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told the instrument is thrown upward from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

$y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}$ (1)

Where:

$y$ is the instrument's final position

$y_{o}=0$ is the instrument's initial position

$V_{o}=15m/s$ is the instrument's initial velocity

$t$ is the time the parabolic movement lasts

$g=2.5\frac{m}{s^{2}}$ is the acceleration due to gravity at the surface of planet X.

As we know $y_{o}=0$ and $y=0$ when the object hits the ground, equation (1) is rewritten as:

$0=V_{o}.t-\frac{1}{2}g.t^{2}$ (2)

Finding $t$ :

$0=t(V_{o}-\frac{1}{2}g.t^{2})$ (3)

$t=\frac{2V_{o}}{g}$ (4)

$t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}$ (5)

Finally:

$t=12s$

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3 years ago

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AysviL [449]

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