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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

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A projectile has an initial x-velocity of 4 m/s, and an initial y-velocity of 27.7 m/s. What is the range of the projectile
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Answer:

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4 0
3 years ago
A tennis ball is dropped from 1.3 m above
Dominik [7]

Answer:

2.65m/s

Explanation:

Using the equation of motion:

v² = u²+2a∆S where

v is the final velocity

u is the initial velocity

∆S is the change in distance

a is the acceleration

Given

u = 0m/s

a = 9.8m/s²

∆S = 1.3-0.943

∆S = 0.357m

Substituting the given parameters into the formula

v² = 0²+2(9.8)(0.357)

v² = 0+6.9972

v² = 6.9972

v=√6.9972

v = 2.65m/s

Hence the velocity at which it hit the ground is 2.65m/s

4 0
3 years ago
Hydrogen is the second most abundant gas in the atmosphere? True or false?
Alenkinab [10]
False as oxygen is the second most abundant and nitrogen is the most abundant at 78%.
7 0
3 years ago
An object that is initially not rotating has a constant torque of 3.6 N⋅m applied to it. The object has a moment of inertia of 6
Korolek [52]

Answer:

0.6

Explanation:

Angular acceleration is equal to Net Torque divided by rotational inertia, which is the rotational equivalent to Newton’s 2nd Law.  Therefore, angular acceleration is equal to 3.6/6 which is 0.6. Hope this helped!

3 0
2 years ago
Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t
n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

y=\frac{\lambda (m+\frac{1}{2})D}{d}

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

\lambda is the wavelength of the light used

In this problem we have:

\lambda=683 nm = 683\cdot 10^{-9} m is the wavelength of the light

d=1.1 mm = 0.0011 m is the width of the slit

m = 13 is the order of the minimum

y=8.57 cm = 0.0857 m is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m

6 0
3 years ago
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