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4 years ago

Why were Aristotle's and Ptolemy's models were accepted for more than a thousand years in spite of being completely incorrect?

Physics

1 answer:

olganol [36]4 years ago

5 0

Answer:

Because it was a major contribution.

Explanation:

At the time people didn't really know any better, and they believed whatever people told them.

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If the horse lifts a 50 kilogram mass a vertical distance of 3 meters in 2 seconds, what is the horse’s minimum power output?

yaroslaw [1]

Answer:

735watts

Explanation:

Power = Force*distance/time

Given

Force = mg = 50 * 9.8'

Force = 490N

distance = 3m

time = 2s

Substitute

Power = 490*3/2

Power = 1470/2

Power = 735Watts

Hence the minimum power output is 735watts

7 0

3 years ago

Does wind have atoms and molecules in int

satela [25.4K]

All forms of matter are comprised of atoms. Wind in itself is the movement of particles(air). Wind does not have particles since it is an action and not a form of matter.

4 0

4 years ago

What is one limitation of using renewable

olga2289 [7]

The answer to this question to A. The other choices are positive.

3 0

3 years ago

What's the formula to find out power

Julli [10]

In general,

   Power = (energy moved) / (time to move the energy) .

If it's mechanical power, then

   Power = (work done) / (time to do the work) .

If it's electrical power, then it can be any one of these:

   Power = (volts) x (amperes)

   Power = (volts)² / (resistance, ohms)

   Power = (amperes)² x (resistance, ohms) .

Whatever kind of energy you're dealing with, power always
turns out to be

(amount of energy produced, used, or moved)
divided by
(time taken to produce, use, or move the energy) .

3 0

3 years ago

the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change

defon

The amount of heat needed to increase the temperature of a substance by $\Delta T$ is given by
$Q=mC_s \Delta T$
where
m is the mass of the substance
$C_s$ the specific heat capacity
$\Delta T$ the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is $C_s = 4.2 J/g ^{\circ}C$ and the amount of heat supplied is $Q=31500 J$ , so if we re-arrange the previous formula we find the increase in temperature of the water:
$\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C$

7 0

3 years ago