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stellarik [79]
3 years ago
14

All of the planets stay near the ecliptic as they move on the Celestial Sphere

Physics
1 answer:
stellarik [79]3 years ago
3 0

Answer:

True

Explanation:

As the Earth goes around the Sun, it will appear that Earth is stationary and Sun is going around it. One can observe the same in real life as well. This apparent path followed by the Sun is called Ecliptic. The plane consisting Ecliptic is called as Ecliptic plane which is same as the orbital plane of Earth.

All the planets of the Solar system are also going around the Sun. Their orbital plane has negligible tilt with respect to Ecliptic plane. Due to this the planets will always appear near to the Ecliptic as they move on the celestial sphere.

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A child looks at his reflection in a spherical Christmas tree ornament 8.0 cm in diameter in season that the image of his face i
malfutka [58]

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1 year ago
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

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For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

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Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

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∴ 1/f1 = 1/(-3.5) + 0

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8 0
3 years ago
An automobile traveling 89.0 km/h has tires of 62.0 cm diameter. (a) What is the angular speed of the tires about their axles? (
bekas [8.4K]

Answer:

a) 79.7rad/s

b) -18.7rad/s^2

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in order to obtain the angular acceleration we have to apply the following formula:

(\omega_f)^2=(\omega_o)^2+2\alpha*\theta\\\\\alpha=-\frac{(\omega_o)^2}{2*rev*2\pi}\\\\\alpha=-\frac{(79.7m/s)^2}{2*27*2\pi}=-18.7rad/s^2

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2 years ago
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