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Ymorist [56]
3 years ago
13

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

Physics
2 answers:
Snezhnost [94]3 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]3 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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Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co
aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

The angular speed of the motion is 2.21433 rad/s

4 0
3 years ago
g stAn experienced spear fisherman sees a small fish swimming in a tidal pool. If the fisherman sees the fish at approximately a
Luba_88 [7]

Answer:

  θ = 28.9

Explanation:

For this exercise let's use the law of refraction

          n₁ sin θ₁ = n₂ sin θ₂

where we use index 1 for air and index 2 for water where the fish is

        sin θ₂ = n₁ / n₂ sin θ₁

in this case the air repair index is 1 and the water 1.33

we substitute

        sin θ₂ = 1 / 1.33 sin t 40

        sin θ = 0.4833

         

        θ = sin⁻¹ 0.4833

        θ = 28.9

5 0
3 years ago
Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal
Step2247 [10]

Answer:

T = 1.2 s

T = 15.1 m = 15 m

Explanation:

This is a case of projectile motion:

TOTAL TIME OF FLIGHT:

The formula for total time of flight in projectile motion is:

T = 2 V₀ Sinθ/g

where,

T = Total Time of Flight = ?

V₀ = Launch Speed = 13.9 m/s

θ = Launch Angle = 25°

g = 9.8 m/s²

Therefore,

T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)

<u>T = 1.2 s</u>

<u></u>

RANGE OF BALL:

The formula for range in projectile motion is:

R = V₀² Sin2θ/g

where,

R = Horizontal Distance Covered by ball = ?

Therefore,

T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)

<u>T = 15.1 m = 15 m</u>

4 0
3 years ago
Read 2 more answers
If a golf ball and a ping-pong ball both move with the same kinetic energy, which has the greater speed? 1. the golf ball 2. can
schepotkina [342]
Think about it like this, the more mass there is, the faster its going to go. If you took a golf ball and a ping pong ball and you held them each separately, you would notice that the golf ball is heavier. If they move with the same kinetic energy, but the golf ball WEIGHS more, then the golf ball will have the greater speed. If you think about it, the ping pong ball may be taking its time to get to wherever its going. 
4 0
4 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug
telo118 [61]

Answer:

a) V_{x}=3.72m/s, b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

0=y_{0}+\frac{1}{2}a_{y}t^{2}

so we can solve for t, so we get:

t=\sqrt{\frac{-(2)y_{0}}{a}}

so now we can substitute the known values, so we get:

t=\sqrt{\frac{-(2)(1.42)}{-9.8}}

which yields:

t=0.538s

So we can use this value to find the velocity in x:

V_{x}=\frac{x}{t}

When substituting we get:

V_{x}=\frac{2m}{0.538s}

which yields:

V_{x}=3.72m/s

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}

We know that V_{0} is zero, so we can simplify the expression:

\Delta y=\frac{V_{yf}^{2}}{2a}

So we can solve the equation for V_{yf}^{2} so we get:

V_{yf}=\sqrt{2\Delta y a}

and when substituting the known values we get:

V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}

which yields:

V_{yf}=-5.28m/s

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

tan \theta =\frac{V_{y}}{V_{x}}

when solving for theta we get:

\theta = tan^{-1}(\frac{V_{y}}{V_{x}})

We can substitute so we get:

\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})

which yields:

\theta = -54.83^{o}

7 0
3 years ago
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