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REY [17]
4 years ago
14

A heavenly body in another's shadow is ____

Physics
2 answers:
Alex_Xolod [135]4 years ago
8 0
<span>It is called a lunar eclipse</span>
krek1111 [17]4 years ago
3 0
It is called a Lunar Eclipse
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What is the most violent star death called
gogolik [260]
Hi there!

The most violent star death is a Supernova. This is the massive explosion of a supergiant star as it's fuel source runs out and it can no longer fuse iron at its core safely. This causes the star to swell to unstable sizes until it explodes in a Supernova. The amount of energy that is output during a supernova is equivalent to all the energy our own Sun outputs in its whole lifetime. 
6 0
4 years ago
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with asecond particle, whi
Anna11 [10]

Answer:

Explanation:

We shall apply conservation of momentum along x and y axis.

Let the final momentum of second particle be p₁ along x axis and p₂ along y axis.

Considering momentum along x axis

2 + 0 = 3 cos 45 + p₁

p₁ = 2-2.12 = - 0.12 kg m/s

Considering momentum along y axis

4 + 0 = 3 sin 45 + p₂

p₂ = 4-2.12 = 1.88  kg m/s

Final momentum = √ ( p₁² + p₂² )

=√ ( .12² + 1.88² )

= 1.88 approx

5 0
3 years ago
During the time a compact disc (CD) accelerates from rest to a constant rotational speed of 477 rev/min, it rotates through an a
atroni [7]

Answer:

<em>126.01 rad/s^2</em>

<em></em>

Explanation:

since it starts from rest, initial angular speed ω' = 0 rad/s

angular speed N = 477 rev/min

angular speed in rad/s ω = \frac{2\pi N}{60} =  \frac{2*3.142* 477}{60} = 49.95 rad/s

angular displacement ∅ = 1.5758 rev

angular displacement in rad/s = 2\pi N = 2 x 3.142 x 1.5758 = 9.9 rad

angular acceleration \alpha = ?

using the equation of angular motion

ω^2 = ω'^2 + 2\alpha∅

imputing values, we have

49.95^{2}  = 0^{2}  + (2 *\alpha*9.9 )

2495 = 19.8\alpha

\alpha = 2495/19.8 = <em>126.01 rad/s^2</em>

8 0
3 years ago
Please answer this question given in the picture
elena-s [515]

A plane mirror forms a virtual image behind the mirror. The image is as far behind the mirror as the object is in front of it. A cannot see his image because the length of the mirror is too short on his side. However, he can see the objects placed at points P and Q, but cannot see the object placed at point R



Hope this helps Buddy!


~ Courtney

6 0
4 years ago
If f3=0 and f1=12n, what does the magnitude of f⃗ 2 have to be for there to be rotational equilibrium?
QveST [7]

<span>I found out F2 and its correct just need help in solving the other parts </span>
<span>F2 = 4.5 N </span>
4 0
4 years ago
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