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2 years ago

Why was the Roman empire able to prosper

Physics

2 answers:

gavmur [86]2 years ago

4 0

The climate of the area. hope this helps! :)

gayaneshka [121]2 years ago

3 0

Answer:

The mild climate enabled Romans to grow wheat, grapes, and olives. This abundance o food supported the people and allowed Rome to prosper. While the climate made year-long agriculture possible, Rome also had the advantage to be near water. The Tiber River helped the agricultural system to prosper

Explanation:

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C. It is radiation leftover from the Big Bang

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2 years ago

Describe how the three methods of thermal energy transfer may take place within the iguana’s enclosure.

sasho [114]

Answer:

Heat can travel from one place to another in three ways: Conduction, Convection and Radiation. Both conduction and convection require matter to transfer heat. Conduction is the transfer of heat between substances that are in direct contact with each other. Thermal energy is transferred from hot places to cold places by convection. Radiation is a method of heat transfer that does not rely upon any contact between the heat source and the heated object as is the case with conduction and convection. Heat can be transmitted through empty space by thermal radiation often called infrared radiation.

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2 years ago

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Bro i need help omg.

sergey [27]

Answer:

25.

Explanation:

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2 years ago

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iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig

Alika [10]

Answer:

453 gm

Explanation:

<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm in air

8 0

1 year ago

water has an index of refraction of 1.33. What is the critical angle for light leaving a pool of water into air? 0 37 90 49

vesna_86 [32]

When light travels from a medium with higher refractive index into a medium with lower refractive index, there is a maximum angle (called critical angle) for which all the light is reflected, so there is no refraction.

The value of the critical angle is given by:
$\theta = \arcsin ( \frac{n_2}{n_1} )$
when n1 is the refractive index of the first medium, while n2 is the refractive index of the second medium. In our case, n1=1.33 (the water) and n1=1.00 (the air). Putting numbers in, we get
$\theta = \arcsin ( \frac{1.00}{1.33} )=49^{\circ}$

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3 years ago