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Charra [1.4K]
3 years ago
6

Why do astronomers use the word on to describe angles on the sky rather than angles in the sky?

Physics
1 answer:
Sever21 [200]3 years ago
8 0
<span>The use of the word on instead of the word in when referring to the angular distance between celestial objects comes about because all of the objects appear to be on the celestial sphere and at an indeterminable distance. While we know that objects are at different distances in the sky, their distance from Earth is irrelevant in determining the angular distance between the two objects as viewed from Earth.</span>
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A weightlifter lifts a 1,250 N barbell 2 m in 3 s. How much power was used to lift the barbell?
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When waves superimpose and make bigger amplitudes what form of interference is that
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Answer:

Constructive Interference

Explanation:

Constructive Interference occurs when two waves superimpose and make bigger amplitudes.

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For Example, In the attachment, two waves A and B superimpose and demonstrate Constructive interference to create the wave C.

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Read 2 more answers
A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at
yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)

Therefore, the work done by the worker in lifting the bucket is given as:

W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)

Now, plug in the values given and solve for 'W'. This gives,

W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0
3 years ago
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