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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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PLS HELP I HAVE EXAM ILL GIVE U BRAINLIST
stiv31 [10]

Explanation:

False

electron and proton attract each other

3 0
3 years ago
A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its
sdas [7]
<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Here the ball travels 3 m less distance in fifth second compared to third second.

That is

           s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² -  u x 2 - 0.5 x g x 2²

           s₃ = u - 2.5 g

Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² -  u x 4 - 0.5 x g x 4²

           s₅ = u - 4.5 g    

That is

           u - 2.5 g = u - 4.5 g + 3

             2 g = 3

                g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

8 0
2 years ago
In the Bohr model of the hydrogen atom, an electron({rm mass};m=9.1; times 10^{ - 31;}{rm kg}) orbits a proton at a distance of
max2010maxim [7]

Answer:

n=6.56×10¹⁵Hz

Explanation:

Given Data

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

To find

Revolutions per second

Solution

Let F be the force of attraction

let n  be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by

F=mω²r......................where ω=2π  n

F=m4π²n²r...............eq(i)

as the values given where

Mass=9.1×10⁻³¹ kg

Radius distance=5.3×10⁻¹¹m

Electric Force=8.2×10⁻⁸N

we have to find n from eq(i)

n²=F/(m4π²r)

n^{2} =\frac{8.2*10^{-8} }{9.11*10^{-31}* 4\pi^{2} *5.3*10^{-11}  }\\ n^{2}=4.31*10^{31}\\ n=\sqrt{4.31*10^{31}}\\ n=6.56*10^{15}Hz

8 0
3 years ago
Ross training for a half marathon which is 13.1 miles long. Within four months he has progressed to an 11 mile practice run if R
damaskus [11]

If Ross takes two months off from training, his fitness level will reduce in comparison to what it was two months ago.

  • In as little as 3–4 weeks after beginning strength training, Ross will probably experience weight increase, energy loss, diminished balance, diminished strength (making it tougher to carry out daily tasks), and overall fewer fitness levels.
  • Many people mistakenly believe they lose muscle mass far more quickly than they actually do because their muscles' ability to store water and glycogen is declining.
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learn more about fitness here: brainly.com/question/13490156

#SPJ10

4 0
2 years ago
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
klio [65]

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:

I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}

8 0
2 years ago
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