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3 years ago

The model of the universe that suggests that the sun is the center of the universe was first brought by

2 answers:

5 0

I think the correct answer from the choices listed above is option D. The model of the universe that suggests that the sun is the center of the universe was first brought by Copernicus. His model is known as the "Sun centered model".

arlik [135]3 years ago

4 0

Copernicus is the right answer. On page 125 of your study material explains that Copernicus put a rotating earth in a sun-centered model. The rotation of earth was able to account for the rising and setting of stars. The orbital motion of the earth and moon also accounted for the motion of the sun and moon with respect to the stars. This was easier to understand but encountered scrutiny due to its differences from religious teachings.

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The isotope of an atom containing 31 protons and 39 neutrons suddenly has two neutrons added to it. What isotope is created?

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It's the second one I hope I'm right

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3 years ago

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Number of waves that pass a given point in one second

Studentka2010 [4]

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3 years ago

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lora16 [44]

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2 years ago

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago

How long did the trip from camp wood to the pacific ocean and back again take?

maria [59]

The trip from Camp Wood to the Pacific Ocean and back again took 1.5 years to complete.

The Lewis and Clark Expedition from May 1804 to September 1806, also known as the Corps of Discovery Expedition, was the first American expedition to cross what is now the western portion of the United States.

6 0

3 years ago