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ycow [4]
3 years ago
11

Why do stars twinkle​

Physics
1 answer:
vlada-n [284]3 years ago
4 0

As light from a star races through our atmosphere, it bounces and bumps through the different layers, bending the light before you see it. Since the hot and cold layers of air keep moving, the bending of the light changes too, which causes the star's appearance to wobble or twinkle.

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The electric potential at a certain distance from a point charge can be represented by V. What is the value of the electric pote
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Answer:

<em>b. At twice the distance, the electric potential is V/2</em>

Explanation:

Electric potential

Is the amount of work needed to move a unit charge from a reference point (usually a point at infinity) without producing acceleration.

The electric potential due to a point charge q at a distance r is given by

\displaystyle V=K\frac{q}{r}

Where K is the Coulomb's constant. If we know the electric potential at a certain distance is V, if the distance is changed to 2r, then the new potential is

\displaystyle V'=K\frac{q}{2r}=\frac{1}{2}K\frac{q}{r}=\frac{1}{2}V

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On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou
BabaBlast [244]

Answer:

<em>176.38 rpm</em>

<em></em>

Explanation:

mass percentage of arms and legs = 13%

mass percentage of legs and trunk = 80%

mass percentage of head = 7%

Total mass of the skater = 74.0 kg

length of arms = 70 cm = 0.7 m

height of skater = 1.8 m

diameter of trunk = 35 cm = 0.35 m

Initial angular momentum = 68 rpm

<em>We assume:</em>

  1. <em>The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.</em>
  2. <em>friction between the skater and the ice is negligible.</em>

We split her body into two systems, the spinning hands as spinning rods

1. Each rod has moment of inertia = \frac{1}{3} mL^{2}

mass m of the arms is 13% of 74 kg = 0.13 x 74 = 9.62 kg

mass of each side will be assumed to be 9.62/2 = 4.81 kg

L = length of each arm

therefore,

I =  \frac{1}{3} x 4.81 x 0.7^{2} = 0.79 kg-m   for each arm

2. Her body as a cylinder has moment of inertia =  \frac{1}{2} mr^{2}

r = radius of her body = diameter/2 = 0.35/2 = 0.175 m

mass of body trunk = (80% + 7%) of 74 kg = 0.87 x 74 = 64.38 kg

I = \frac{1}{2} x 64.38 x 0.175^{2} = 0.99 kg-m

We consider each case

case 1: Body spinning with arm outstretched

<em>Total moment of inertia = sum of moments of inertia of both arms and moment of inertia of body trunk</em>

I = (0.79 x 2) +  0.99 = 2.57 kg-m

angular momentum = Iω

where ω = angular speed = 68.0 rpm = \frac{2\pi }{60} x 68 = 7.12 rad/s

angular momentum = 2.57 x 7.12 = 18.29 kg-rad/m-s

case 2: Arms pulled down parallel to trunk

<em>The momentum of inertia will be due to her body trunk alone</em> which is 0.91 kg-m

angular momentum =  Iω

=  0.99 x ω = 0.91ω

<em>according to conservation of angular momentum, both angular momentum must be equal</em>, therefore,

18.29 = 0.99ω

ω = 18.29/0.99 = 18.47 rad/s

18.47 ÷ \frac{2\pi }{60}  = <em>176.38 rpm</em>

7 0
3 years ago
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