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iren [92.7K]
3 years ago
10

Over a period of one year, how much of the overall (night) sky would a Bellingham observer be able to see?

Physics
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

The overall sky a Bellingham observer would see is about 75-100% over the period of 1 year.

Explanation:

Due to following factors, the overall sky visibility will fall in this range.

  1. The location of Bellingham on Earth is in the Northern hemisphere over the equator.
  2. The tilt of the earth cannot allow the visibility of the complete sky.
  3. The light pollution and aerosol concentrations has reduced the sky visibility at night.
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Estimate how much solar energy reaches the earth per year (in Joule).
Alexxandr [17]

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3 years ago
two porters are available to carry a long timber wood.out of them one is weak.how to do you make less load to the weak one? writ
marshall27 [118]

Answer:

Please find the answer in the explanation.

Explanation:

Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?

We can make the weak one to carry less load through two different ways or means.

First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.

Second, the strong porter can carry the long timbre wood almost to the fulcrum and allow the weak one to support at the other end. By doing this, the weak one will only carry light portion of the load.

4 0
3 years ago
An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.
Ber [7]

Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

Explanation:

Given that,

Mass of first lead ball, m_1=5.2\ kg

Mass of the other lead ball, m_2=0.046\ kg

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N

So, the magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N. Hence, this is the required solution.

5 0
2 years ago
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