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Virty [35]
3 years ago
14

Because the direction of Earth's motion around the Sun continually changes during the year, the apparent position of a star in t

he sky moves in a small loop, known as the aberration of starlight. In order to better understand this phenomenon, it is sometimes helpful to use visual analogies. In these visual analogies, the car is analogous to the Earth, and the rainfall is analogous to starlight. Determine which visual analogies correspond to the following scenarios:
A) The Earth moving around the Sun and interacting with light from a distant star
B) A person on the moving Earth observing the light from a distant star
C) A person on a motionless Earth observing the light from a distant star
Physics
1 answer:
Alenkasestr [34]3 years ago
3 0

Answer:

B) A person on the moving Earth observing the light from a distant star

Explanation:

The aberration of starlight is a <u>phenomenon</u> in <u>Astronomy</u> used to describe the '<u>seeming</u>' or apparent movement of a star around its true position. This seeming' or apparent movement of a star is caused by and dependent on the velocity of the observer. The movement of the observer relative to the star creates the notion that the star is moving. However, in reality, the star is static.  The star is not moving, what is really moving is the observer; this movement of the observer is what makes it seem that the star was moving. It is not an optical illusion, although, the effect seems close enough.

For example, in the analogy given:

A car moving in the rain or under the rainfall

The car is the one <u>moving</u> in the same way that the Earth is the one <u>revolving</u>. The rainfall drops <u>vertically</u> in the same way that the starlight is static. The movement of the car relative to the rainfall is what makes it seem that the rain is falling 'diagonally'. The same visual analogy is observe when a person on the moving Earth observing the light from a distant star

Hence, the correct option is B

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6 0
3 years ago
What are the two factors that determine the strength of the gravitational force between two objects?
Lena [83]

Answer:

mass and distance

Explanation:

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4 0
2 years ago
One ball of mass 0.600kg travelling 9.00m/s to the right collides head on elastically with a second ball of mass 0.300kg travell
Alina [70]

Let m₁ and v₁ denote the mass and initial velocity of the first ball, and m₂ and v₂ the same quantities for the second ball. Momentum is conserved throughout the collision, so

m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'

where v₁' and v₂' are the balls' respective velocities after the collision.

Kinetic energy is also conserved, so

1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁')² + 1/2 m₂ (v₂')²

or

m₁ v₁² + m₂ v₂² = m₁ (v₁')² + m₂ (v₂')²

From the momentum equation, we have

(0.600 kg) (9.00 m/s) + (0.300 kg) (-8.00 m/s) = (0.600 kg) v₁' + (0.300 kg) v₂'

which simplifies to

10.0 m/s = 2 v₁' + v₂'

so that

v₂' = 10.0 m/s - 2 v₁'

From the energy equation, we have

(0.600 kg) (9.00 m/s)² + (0.300 kg) (-8.00 m/s)² = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

which simplifies to

67.8 J = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²

or

226 m²/s² = 2 (v₁')² + (v₂')²

Substituting v₂' yields

226 m²/s² = 2 (v₁')² + (10.0 m/s - 2 v₁')²

which simplifies to

3 (v₁')² - (20.0 m/s) v₁' - 63.0 m²/s² = 0

Solving for v₁' using the quadratic formula gives two solutions,

v₁' ≈ -2.33 m/s   or   v₁' = 9.00 m/s

but the second solution corresponds to the initial conditions, so we omit that one.

Then the second ball has velocity

v₂' = 10.0 m/s - 2 (-2.33 m/s)

v₂' ≈ 14.7 m/s

6 0
2 years ago
A uniform rod is attached to a wall by a hinge at its base. The rod has a mass of 6.5 kg, a length of 1.3 m, is at an angle of 1
liubo4ka [24]

Answer:

tension wire 104 N, horizontal force hinge 104 N, vertical force hinge 63.7 N

Explanation:

The system described is in static equilibrium under the action of several forces, which are shown in the attached diagram, where T is the tension of the wire, W is the weight of the bar, Fx is the horizontal reaction of the wall and Fy It is the vertical reaction of the wall.

The directions of the forces are indicated by the arrows and are marked intuitively, but when solving the problem if one gives a negative value this indicates that the direction is wrong, but does not alter the results of the problem

For the resolution we use Newton's second law, both in translational and rotational equilibrium, if necessary.

We must establish a reference system to assign the positive meaning, we place it with the origin in the hinge, and the positive directions to the right and up. The location of the coordinate system allows us to eliminate the reaction of the hinge by having zero distance to origin. We write the equilibrium equations for each axis

     ∑ Fx = 0

     Fx -T =0

     ∑ Fy =0

     Fy -W =O      W = m g

     ∑τ =0

      -T dy + W dx =0

For rotational equilibrium we take the positive direction as counterclockwise rotation and distance is the perpendicular distance of the force to the axis of the coordinate system

     

       dy = L sin 17

      dx = (L/2) cos 17

Where L is the length of the bar and the weight is applied at the center of it, we write and simplify the equation

    -T L sin 17 + mg (L / 2) cos 17 = 0

   - T sin 17 + mg/2  Cos 17 =0

We write the three equations together

     Fx- T = 0

     Fy -W = 0

     -T sin 17 + (m g / 2) cos 17 = 0

a) With the third equation we can find the wire tension

     T = m g / 2 cos 17 / sin17

     T = 6.5  9.8/2 cotan 17

     T = 104 N

b) We use the first equation to find Fx

       Fx- T =0

       Fx = T = 104 N

c) We use the second equation to find Fy

 

       Fy = W = m g

       Fy = 6.5 9.8

       Fy = 63.7 N

4 0
3 years ago
A centrifuge accelerates uniformly from rest to 15000 rpm in 330 s . Through how many revolutions did it turn in this time?
eduard

Answer:

The number of revolutions turned by the centrifuge is 8250 revolutions.

Explanation:

Given;

number of revolution per minutes, ω = 15000 rpm

time of  motion, t = 330 s = 5.5 minutes

The number of revolutions turned by the centrifuge is given by;

N = \frac{1500 \ Rev}{minutes} *5.5 \ minutes\\\\N = 8250 \ revolutions

Therefore, the number of revolutions turned by the centrifuge is 8250 revolutions.

4 0
3 years ago
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