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solong [7]
3 years ago
12

The distance to the North Star, Polaris, is approximately 6.44 3 1018 m.

Physics
1 answer:
zhenek [66]3 years ago
8 0

Answer:

a) 2,042.1 years. b) 500 sec. c) 2.56 sec.

Explanation:

Visible light, is just an electromagnetic wave approximately within 400-700 nm) , so , in vacuum, propagates at a constant speed of 3. 10⁸ m/s.

So, in order to get the time needed for the light coming from any point of the universe can reach us, we just need to apply the velocity definition equation, as follows:

v = Δx / Δt, and solve for Δt, taking into account units.

a) if d = 6.44. 3.10¹⁸ m, and v = 3.10⁸ m/s, we find t as follows:

t = 6.44.3. 10¹⁸ / 3. 10⁸ s = 6.44 . 10¹⁰ sec.

Converting seconds to years:

6.44. 10¹⁰ sec . (1min/60 sec).(1h/60 min). (1d/24h)(1y/365d) = 2,042.1 years.

b) As the average distance earth-sun is 1.5.10⁸ m, we find t as follows:

t = 1.5 10¹¹ m / 3.10⁸ m/s = 500 sec = 8.33 min

c) As the average distance Earth-Moon is 3.84. 10⁸ m, we find t to be as follows:

t = 2. (3.84. 10⁸ m / 3. 10⁸ m/s) = 2.56 sec.

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A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 5100 MJ with respect to Eart
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To solve this problem we will apply the theorem given in the conservation of energy, by which we have that it is conserved and that in terms of potential and kinetic energy, in their initial moment they must be equal to the final potential and kinetic energy. This is,

E_{initial} = E_{final}

PE_{initial}+KE_{initial} = PE_{final}+KE_{final}

Replacing the 5100MJ for satellite as initial potential energy, 4200MJ for initial kinetic energy and 5700MJ for final potential energy we have that

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As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis
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Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

d = v_0t + \frac{1}{2}at^2

The block starts from rest, so:

d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2

Now, we can do a summation of forces of the block using Newton's Second Law:

F = ma = m_bg - T

mb = mass of the block

T = tension of string

Solve for tension:

T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N

Now, we can do a summation of torques for the wheel:

\Sigma \tau = rF\\\\\Sigma\tau = rT

Rewrite:

I\alpha = rT

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2

Now, plug in the values into the equation:

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