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2 years ago

Select all that apply. Which of the following astronomers supported the Earth-centered system?

Physics

2 answers:

Vilka [71]2 years ago

8 0

Answer:

Ptolemy and partially Tycho Brahe

Explanation:

Both Ptolemy and Tycho Brahe believed that the earth was the center of the orbit of the sun, the first also of all the other planets and stars, the later believed the sun was in exchange the center of all the other celestial bodies, then Brahe's posture was an intermediary between that of Ptolemy and the heliocentrism(Sun as center of the solar system, where the planets orbit around it) of Copernicus, that proved to be the right observation later with galileo galilei.

Stella [2.4K]2 years ago

6 0

Among the options, the two astronomers who supported the Earth-centered system were Tycho Brahe and Ptolemy.

Both astronomers developed in fact a system where the Sun was located at the center, while the Earth and the other planets were orbiting around the Sun.

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Answer:

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Explanation:

Lets calculate

The weight attached to the spring is 24 pounds

Acceleration due to gravity is $32ft/s^2$

Assume x , is spring stretched length is ,4 inches

Converting the length inches into feet $x=\frac{4}{12} =\frac{1}{3}feet$

The weight (W=mg) is balanced by restoring force ks at equilibrium position

mg=kx

$W=kx$ ⇒ $k=\frac{W}{x}$

The spring constant , $k=\frac{24}{1/3}$

= 72

If the mass is displaced from its equilibrium position by an amount x, then the differential equation is

$m\frac{d^2x}{dt} +kx=0$

$\frac{3}{4} \frac{d^2x}{dt} +72x=0$

$\frac{d^2x}{dt} +96x=0$

Auxiliary equation is, $m^2+96=0$

$m=\sqrt{-96}$

= $\frac{+}{} i4\sqrt{6}$

Thus , the solution is $x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}$

$x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6t}$

The mass is released from the rest x'(0) = 0

$=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2$ $4\sqrt{6}$ $cos4\sqrt{6(0)}$ =0

$c_2$ $4\sqrt{6} =0$

$c_2=0$

Therefore , $x(t)=c_1$ $cos 4\sqrt{6t}$

Since , the mass is released from the rest from 4 inches

$x(0)= -4$ inches

$c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}$ feet

$c_1=-\frac{1}{3}$ feet

Therefore , the equation of motion is $-\frac{1}{3} cos4\sqrt{6t}$

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A wheelhas a radius of 4.8 m. Howfar (path length) does a point on the circumference travel if thewheel is rotated through angle

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Answer:

(a) 2.512 m

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As we know that in 360 degree it rotates a complete round that means circumference.

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so,

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