Which of these assumptions is Ptolemy's model of planetary motion based on?

2 answers:

4 0

Ptolemy's model of planetary motion based on the fact that The sun moves around the earth.
Ptolemy thought that all celestial objects — including the planets, Sun, Moon, and stars — orbited Earth. Earth, in the center of the universe, did not move at all.

djverab [1.8K]3 years ago

3 0

Ptolemy's model of planetary motion is based on that the sun moves around the earth.

You might be interested in

Which is not a property of a pure substance

Ann [662]

air is a mixture and not a pure substance.

5 0

3 years ago

HELP ITS DUE IN 4 MINUTES

DochEvi [55]

Answer:

The nephew

Explanation:

The nephew has a seeing eye dog which means the nephews tracks are the dogs tracks, and the dog tracks are not covered by anything which means the dog was last and stole the butterfly.

<u><em>Hope this helps!</em></u>

6 0

3 years ago

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

$p_{ox} = p_{oAx} + p_{oBx} (1)$

We can do exactly the same for the initial momentum along the y-axis:

$p_{oy} = p_{oAy} + p_{oBy} (2)$

The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

$p_{fx} = (m_{A} + m_{B} ) * v_{fx} (3)$

We can repeat the process for the y-axis, as follows:

$p_{fy} = (m_{A} + m_{B} ) * v_{fy} (4)$

Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

$v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} = 2.96 m/s (5)$

In the same way, we can find the component of the final momentum along the y-axis, as follows:

$v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} = 3.15 m/s (6)$

With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

$v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)$

The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

$p_{f} = (m_{A} + m_{B})* v_{f} = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)$

Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
We can find this angle applying the definition of tangent of an angle, as follows:

$tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)$