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Stolb23 [73]
4 years ago
9

Why were epicycles necessary in ptolemy’s model of the universe?

Physics
2 answers:
N76 [4]4 years ago
4 0

Because Mars, Jupiter, and Saturn each appear to slow down, stop, move backwards for a while, then again slow down, stop, and start moving forward again. Each plant does this at some time during each year.

If they were simply moving in circles around the Earth, they couldn't do any of this. They would need to be moving on small circles attached to the big circles. The small ones are the 'epicycles'.

Liono4ka [1.6K]4 years ago
3 0

epicycles were orbits within orbits used to explain discrepancies between expected and observed planetary movement, including the appearance of planets slowing down, speeding up, and moving backward.

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C) light waves travel faster than sound waves

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4 years ago
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A meter stick is supported by a pivot at its center of mass. Assume that the meter stick is uniform and that the center of mass
iragen [17]

Answer:

64.5 cm

Explanation:

30 * 80 + x * 110 = 50* 190 => x = 64.5

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3 years ago
As a diligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.33 m l
deff fn [24]

Answer: 62 μT

Explanation:

Given

Length of rod, l = 1.33 m

Velocity of rod, v = 3.19 m/s

Induced emf, e = 0.263*10^-3 V

Using Faraday's law, the induced emf of a rod can be gotten by the formula

e = blv where,

e = induced emf of the rod

b = magnetic field of the rod

l = length of the rod

v = velocity of the rod. On substituting, we have

0.263*10^-3 = b * 1.33 * 3.19

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8 0
3 years ago
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Four forces are exerted on a disk of radius R that is free to spin about its center, as shown above. The magnitudes are proporti
Dmitry_Shevchenko [17]

The given magnitude of forces of F₁ = F₄, F₂ = F₃, F₁ = 2·F₂, give the

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<h3>How can the net torque on the disk be calculated?</h3>

The given parameters are;

F₁ = F₄

F₂ = F₃

F₁ = 2·F₂

Therefore;

F₄ = 2·F₂

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\vec{F_4} = \mathbf{\frac{\sqrt{3} }{2} \cdot F_4 \cdot \hat i -  0.5 \cdot F_4 \hat j}

\vec{F_2} = \mathbf{ -F_2 \,  \hat j}

Clockwise moment due to F₄, M₁ = -0.5 \times F_4 \,  \hat j  \times \dfrac{R}{2}

Therefore;

M_1  =- 0.5 \times 2 \times  F_2 \,  \hat j  \times \dfrac{R}{2} =   \mathbf{ -F_2 \,  \hat j  \times \dfrac{R}{2}}

Counterclockwise moment due to F₂ = -F_2 \,  \hat j  \times \dfrac{R}{2}

Given that the clockwise moment due to F₄ = The counterclockwise moment due to F₂, we have;

Two forces that combine to exert zero net torque on the disk are;

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Learn more about the resolution of vectors here:

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2 years ago
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3 years ago
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