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Svetllana [295]
3 years ago
6

Is there an eclipse happening soon in England

Physics
2 answers:
lyudmila [28]3 years ago
3 0

A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.

England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.

For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge.  It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM.  Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.

The catch in observing the eclipse is:

                       <em><u>YOU MUST NOT LOOK AT THE SUN</u></em>.

Staring at the sun for a period of time can cause permanent damage to
your vision, even though <em><u>you don't feel it while it's happening</u></em>.

This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours.  Please look
in your local newspaper, or search online for phrases like "safe eclipse
viewing".


Alja [10]3 years ago
3 0
Yes, on the 20th of March. <span>The eclipse will begin at 8.45am and the Moon will be nearest the middle of the Sun at 9.31am. </span>
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A steel pipe of 12-in. outer diameter is fabricated from 1 4 -in.-thick plate by welding along a helix that forms an angle of 25
Varvara68 [4.7K]

Answer:

Normal stress = 66/62.84 = 1.05kips/in²

shearing stress  = T/2 = 0.952/2 = 0.476 kips/in²

Explanation:

A steel pipe of 12-in. outer diameter  d₂ =12in  d₁= 12 -4in = 8in

4 -in.-thick  

angle of 25°

Axial force P = 66 kip axial force

determine the normal and shearing stresses

Normal stress б = force/area = P/A

           = 66/ (П* (d₂²-d₁²)/4

           =66/ (3.142* (12²-8²)/4

          = 66/62.84 = 1.05kips/in²

Tangential stress T = force* cos ∅/area = P/A

           = 66* cos 25/ (П* (d₂²-d₁²)/4

           =59.82/ (3.142* (12²-8²)/4

          = 59.82/62.84 = 0.952kips/in²

shearing stress  = tangential stress /2

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3 years ago
The light from Star 1 reaches Star 3 in 138 years and Star 2 in 63 years. Which of these conclusions about the stars is correct?
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To solve this exercise, it is necessary to apply the concepts of conservation of the moment especially in objects that experience an inelastic colposition.

They are expressed as,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where,

m_1= mass of the skier

m_2= mass of the cat

v_1 = initial velocity of skier

v_2 = initial velocity of cat

v_f= final velocity of both

Re-arrange to find V_f we have,

V_f = \frac{m_1v_1+m_2v_2}{(m_1+m_2)}

V_f = \frac{(60)(15)+(5)(-3.8)}{(60+5)}

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Once the final velocity is found it is possible to calculate the change in kinetic energy, so

\Delta KE = KE_i-KE_f

\Delta KE = \frac{1}{2}(m_1v_1^2+m_2v^2_2)-\frac{1}{2}(m_1+m_2)v_f^2

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\Delta KE = 819.1J

Therefore the amount of kinetic energy converted in to internal energy is 819J

3 0
3 years ago
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