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antoniya [11.8K]
3 years ago
8

What is responsible for the apparent motion of the Moon in the sky from East to West?

Physics
1 answer:
DiKsa [7]3 years ago
6 0

As a result, although the Moon is moving to the east relative to the stars, the much faster westward motion of the sky is carrying it to the west, so despite its eastward motion relative to the center of the Earth, it rises in the east and sets in the west, just like any other celestial body.


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The average distance an electron travels between collisions is 2.0 μmμm . What acceleration must an electron have to gain 2.0×10
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The solution is in the attachment

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1.)State which definition of power below is correct.
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A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is mounted o
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The total angular momentum of the system about point B is L=m_1r_1\omega_1+m_2r_2\omega_2

Angular momentum, also known as moment of momentum or rotational momentum, is the rotating counterpart of linear momentum.

A rigid object's angular momentum is defined as the product of its moment of inertia and its angular velocity. If there is no external torque on the object, it is analogous to linear momentum and is subject to the fundamental constraints of the conservation of angular momentum principle. The vector quantity angular momentum It is derived from the expression for a particle's angular momentum.

Given,

mass of ball 1 = m1

m₂ mass of ball 2=m2

v₁ is the velocity of ball=r₁ω₁

v₂ is the velocity of ball 2=r₂ω₂

The total angular momentum is given as;

V_{total}=r_1\omega_1+r_2\omega_2\\\\L=m_1r_1\omega_1+m_2r_2\omega_2

Hence the total angular momentum  will be L=m_1r_1\omega_1+m_2r_2\omega_2

To learn more about angular momentum refer here

brainly.com/question/29512279

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6 0
1 year ago
What is the acceleration of the the object during the first 4 seconds?
AVprozaik [17]

Answer:

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

6 0
2 years ago
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