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4 years ago

Copernicus's model of the universe differs from Ptolemy's because Copernicus believed

Physics

2 answers:

Art [367]4 years ago

5 0

Answer:

Earth rotates on its axis and orbits the sun.

Explanation:

Ptolemy believed earth is stationary and is located at the center of the universe. Every celestial object moves around the earth.

It was easier to assume that the earth is not moving since it appears to be stationary to people dwelling on it and to assume that the celestial objects are actually rotating the earth since they appear to be moving.

This is the perfect example that looks can be deceiving. In the ancient times the circle is believed to be a divine shape since it is perfect and the two major celestial objects, the sun and the moon, moves in a circular path. But it turns out, not everything rotates the earth in circular orbit. So Ptolemy proposed nested circular orbits. Which was so complex and farfetched.

Copernicus, a student of Tycho Brahe, (remember Tycho Brahe is an advocate of geocentric theory) inherited Tycho brahe's records of the movements of mars. He could not explain the retrograde motion of mars within the limits of geocentric theory.

He then suggested to remove the concept of divinity, associated with the earth being the center of all universe and to treat earth like every other celestial object. He was the first ever radical who steered us on the right path. In his heliocentric theory, earth rotates on its axis and orbits the sun. This can explain the retrograde motion of mars very easily

Aneli [31]4 years ago

4 0

Answer:

That the Sun was the center of the universe.

Explanation:

Ptolemy believed in geocentrism, which means that the Earth is the center of the Universe. This was also supported by Aristotle. Copernicus on the other hand believed in a heliocentric system, which proposes that the sun is the center of the universe.

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A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of the E

Sveta_85 [38]

Answer:

995 N

Explanation:

Weight of surface, w= 4000N

Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

m= 4000/9.81= 407.7472

Using radius of orbit of 6371km

The force of gravity of satellite in its orbit, $F=\frac {GMm}{(2r)^{2}}=\frac {GMm}{4(r)^{2}}$

Where $G=6.67*10^{-11}$ and $M=5.94*10^{24}$

$F=\frac {(6.67*10^{-11}*5.94*10^{24}*407.7472)}{4*({6.371*10^{6}m)}^{2}}$

F= 995.01142 then rounded off

F=995N

6 0

3 years ago

During a very quick stop, a car decelerates at 28.4 rad/s?. Assume the tires initially rotated in the positive direction and rad

Damm [24]

Answer:

a) 24

b) 3.3 sec

c) 29.8 m/s

d) 48.85 m

Explanation:

α = angular acceleration = - 28.4 rad/s²

r = radius of the tire = 0.32 m

w₀ = initial angular velocity = 93 rad/s

w = final angular velocity = 0 rad/s

θ = angular displacement

Using the equation

w² = w₀² + 2αθ

0² = 93² + 2 (- 28.4) θ

θ = 152.3 rad

n = number of revolutions

Number of revolutions are given as

$n = \frac{\theta }{2\pi }$

$n = \frac{152.3 }{2(3.14) }$

$n = 24$

t = time taken to stop

using the equation

w = w₀ + αt

0 = 93 + (- 28.4) t

t = 3.3 sec

v₀ = initial velocity of the car

initial velocity of the car is given as

v₀ = r w₀ = (0.32) (93) = 29.8 m/s

v = final velocity = 0 m/s

a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²

d = distance traveled by car before stopping

Using the equation

v² = v₀² + 2 a d

0² = 29.8² + 2 (- 9.09) d

d = 48.85 m

8 0

4 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

<u>The answer for (a) is 1.6m</u>

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

<u>the answer to (b) is 6.86m</u>

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

<u>the answer to (c) is 2.86s</u>

6 0

3 years ago

Complete the following:

masha68 [24]

When light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

To find the answer, we have to know about the rules followed by drawing ray-diagram.

<h3>What are the rules obeyed by light rays?</h3>

If the incident ray is parallel to the principal axis, the refracted ray will pass through the opposite side's focus.
The refracted ray becomes parallel to the major axis if the incident ray passes through the focus.
The refracted ray follows the same path if the incident light passes through the center of the curve.

Thus, we can conclude that, when light is incident parallel to the principal axis and then strikes a lens, the light will refract through the focal point on the opposite side of the lens.

Learn more about refraction by a lens here:

brainly.com/question/13095658

#SPJ1

8 0

2 years ago

An electronic charge can be transmitted through?

DedPeter [7]

Answer:

An electronic charge can be transmitted through the human body, water and metals.

6 0

4 years ago