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Ymorist [56]
4 years ago
13

How do we prove the earth is round? What evidence do we have and reasoning?

Physics
2 answers:
Snezhnost [94]4 years ago
8 0

Just like any other scientific conclusion ("theory"), we DON'T say the conclusion first and then go out and look for evidence that supports it. The "scientific" process is to gather up ALL the evidence we possibly can, and THEN figure out a simple description that can explain the evidence. After that, if our description turns out to predict things that we HAVEN'T seen yet but they do turn out to be true, that gives us more confidence that our description is good, and probably true and accurate.

When you live for several years, and you talk to a lot of people, and there's a lot of discussion of day and night, sunrise and sunset, moonrise and moonset, the looks and motions of all the things in the sky seen from different places at different times, how mountains and clouds look during twilight, how ships look when they're coming in and going out, how distant mountains appear, the idea of the Earth as a spinning sphere is the one that can explain them all. No other description can, unless it's hilariously complicated to start with, and you keep making up new kinks and patches every time somebody finds a problem with it.

This is where we were about 2000 years ago.

Since then, telescopes were invented, and we could see that all the OTHER planets are spinning spheres.

And just yesterday, when we started flinging cameras out, turning them around, and taking far-out pictures of the Earth, guess what ! The pictures show that the Earth is spherical. And when we take videos, we see it spinning.

Our theory is looking better all the time.

Jlenok [28]4 years ago
7 0

Answer: Go to the harbor. When a ship sails off toward the horizon, it doesn't just get smaller and smaller until it's not visible anymore. Instead, the hull seems to sink below the horizon first, then the mast. When ships return from sea, the sequence is reversed: First the mast, then the hull, seem to rise over the horizon.

Climbing to a high point will allow you to be able to see farther if you go higher. If the Earth was flat, you'd be able to see the same distance no matter your elevation

You might be interested in
17. A volleyball weighs about 300 grams.
atroni [7]

Answer:

PE = 44.1 J

Explanation:

Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)

We have:

  • 1 kilograms = 1000 grams.

We convert it using a rule of 3, replacing, simplifying units and solving:

  • \boxed{\bold{x=\frac{gr*1\ kg}{1000\ gr}=\frac{300\ gr*1\ kg}{1000\ gr}=\frac{300\ kg}{1000}=\boxed{\bold{0.3\ kg}}}}

==================================================================

Earth's gravity is known to be 9.8 m/s², so we have:

Data:

  • m = 0.3 kg
  • g = 9.8 m/s²
  • h = 15 m
  • PE = ?

Use formula of potencial energy:

  • \boxed{\bold{PE=m*g*h}}

Replace and solve:

  • \boxed{\bold{PE=0.3\ kg*9.8\frac{m}{s^{2}}*15\ m}}
  • \boxed{\boxed{\bold{PE=44.1\ J}}}

Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.

The potential energy of the volleyball is <u>44.1 Joules.</u>

Greetings.

8 0
3 years ago
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between
erma4kov [3.2K]

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   COP  =  \frac{T_i}{T_o - T_i}

Here T_i is the inside temperature

while  T_o is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   \frac{T_i}{T_o - T_i} heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         Q \ \alpha \ (T_o - T_i)

=>        Q= k (T_o - T_i)

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   \frac{T_i}{T_o - T_i}  amount of heat

   E  unit of electricity will remove  Q= k (T_o - T_i)

So

      E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }

=>   E = \frac{k}{T_i} (T_o - T_i)^2

given that  \frac{k}{T_i} is constant

    =>  E \  \alpha  \  (T_o - T_i)^2

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  T_i   =  30 ^oC

 and      T_o  =  40 ^oC

Hence  

     E = K (T_o - T_i)^2

Here K stand for a constant

So  

        E = K (40 -  30)^2

=>      E = 100K

Now if  the  T_i   =  20 ^oC

Then

       E = K (40 -  20)^2

=>      E = 400 \ K

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       Q = k (T_o - T_i )^{\frac{1}{2} }

So

       E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }

=>   E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }

Assuming \frac{k}{T_i} is a constant

Then  

     E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

   

4 0
3 years ago
an aircraft landing on an air craft carrier is brought to a complete stop from an inital velocity of 215km/hr in 2.7 seconds. wh
worty [1.4K]

u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2



Hope it helps
8 0
3 years ago
Use the Electrostatic Series to predict the charges on the following pairs of substances when they are rubbed together:
Lisa [10]
Answer: Not 100% sure but I believe the answer is B.

Hope this helps! ^^
6 0
3 years ago
From the deepest to the surface, what are the parts of the earth's interior?
shepuryov [24]

The answer is A: Core --> Mantle --> Crust.

Core: The earth's core is the center of the earth, which would ultimately be the deepest. The core is made up of alloy, which is a mixture of many medals, such as iron and nickel.

Mantle: The earth's mantle is the layer between the earths crust and core. Often made of silicate rocks.

Crust: The earth's crust is the outer-most of the three options. Usually made of up different types of rocks.

3 0
3 years ago
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