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3 years ago

14

Which of these discoveries is generally attributed to kepler?

1 answer:

Ne4ueva [31]3 years ago

4 0

The question is incomplete and no options are given but the answer is;
The orbit of Mars is an ellipse,<span> is generally attributed to Kepler.
Kepler's first law states that; "</span><span>The orbits of the planets are ellipses, with the Sun at one focus of the ellipse.
</span>The planet at that point takes after and follow the ellipse in its orbit, which implies that the distance between Earth and Sun remove is continually changing as the planet goes around its orbit.

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The length of the student desk is measured using a

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3 significant figures are in 1.02m

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3 years ago

Read 2 more answers

A satellite with mass 6000 kg is orbiting the planet at 2500 km above the planet's

9966 [12]

By the law of universal gravitation, the gravitational force <em>F</em> between the satellite (mass <em>m</em>) and planet (mass <em>M</em>) is

<em>F</em> = <em>G</em> <em>M</em> <em>m</em> / <em>R </em>²

where

<em>• G</em> = 6.67 × 10⁻¹¹ m³/(kg•s²) is the universal gravitation constant

• <em>R</em> = 2500 km + 5000 km = 7500 km is the distance between the satellite and the center of the planet

Solve for <em>M</em> :

<em>M</em> = <em>F R</em> ² / (<em>G</em> <em>m</em>)

<em>M</em> = ((3 × 10⁴ N) (75 × 10⁵ m)²) / (<em>G</em> (6 × 10³ kg))

<em>M</em> ≈ 2.8 × 10¹⁴ kg

6 0

3 years ago

A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera

AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel $r=0.3\ m$

Initial angular velocity $\omega _0=0$

It rotates 3 revolution in 5 s therefore

$\omega =2\pi 3=\6\pi =18.85\ rad/s$

using $\omega =\omega _0+\alpha t$

where $\alpha =angular\ acceleration$

$\omega =Final\ angular\ velocity$

$t=time$

$\alpha =\frac{18.85}{5}=3.77 rad/s^2$

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

$\omega at t=1$

$\omega =0+3.77\times 1=3.77 rad/s$

$a_c=\omega ^2\cdot r$

$a_c=(3.77)^2\cdot 0.3=4.26 m/s^2$

Tangential acceleration $a_t=\alpha \times r$

$a_t=3.77\times 0.3=1.13 m/s^2$

$a_{net}=\sqrt{a_t^2+a_c^2}$

$a_{net}=\sqrt{(1.13)^2+(4.26)^2}$

$a_{net}=4.41 m/s^2$

7 0

3 years ago

The latitude of any location on earth is the angle formed by the two rays drawn from the center of earth to the location and to

Alina [70]

The distance between city a and city b is 833.345 miles.

We know that

1°=60'

The distance of city a from the initial ray is calculated as

x_a=3960*tan45.46°=4024.101 miles

The distance of city b from the initial ray is calculated as

x_b=3960*tan 38.86°=3190.75 miles

Now the distance between city a and b is equal to

4024.101-3190.75=833.345 miles

This is the vertical distance between the cities.

5 0

3 years ago