A ___ is one way a star might die and it can trigger the beginning of a new stars life cycle .

Physics

1 answer:

Goryan [66]4 years ago

4 0

The correct answer is Supernova.

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A rocket ship leaves Earth's atmosphere. Its initial velocity is less than its final velocity. What is this an example of?

77julia77 [94]

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3 years ago

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Our galaxy is Select one: a. one member of a large cluster of thousands of galaxies. b. one member of a small cluster of galaxie

Art [367]

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3 years ago

PLEASE SOLVE QUICKLY!!!<br> Solve for A, B, and C from graph

hram777 [196]

A = 59.35cm

B = 196.56g

C = 74.65g

<u>Explanation:</u>

We know,

$x = \frac{L}{\frac{W}{F} +1}$

and L = x+y

Total length, L = 100cm

Weight of Beam, W = 71.8g

Center of mass, x = 49.2cm

Added weight, F = 240g

Position weight placed from fulcrum, y = ?

$L-y = \frac{L}{\frac{W}{F}+1 } \\100 - y = \frac{100}{\frac{71.8}{49.2}+1 } \\100 - y = \frac{100}{1.46+1}\\\\100 - y = \frac{100}{2.46} \\100-y = 40.65\\\\y = 59.35cm$

Therefore, position weight placed from fulcrum is 59.35cm

Total length, L = 100cm

Center of mass, x = 47.8 cm

Added weight, F = 180g

Position weight placed from fulcrum, y = 12.4cm

Weight of Beam, W = ?

$47.8 = \frac{100}{\frac{W}{180}+1 }\\\47.8 = \frac{100}{\frac{W+180}{180} } \\\\47.8 = \frac{100 X 180}{W+180}\\ \\47.8W + 47.8 X 180 = 18000\\47.8W = 18000 - 8604\\W = \frac{9396}{47.8}\\ W = 196.56g$

Therefore, weight of the beam is 196.56g

Total length, L = 100cm

Center of mass, x = 50.8 cm

Position weight placed from fulcrum, y = 9.8cm

Weight of Beam, W = 72.3g

Added weight, F = ?

$50.8 = \frac{100}{\frac{72.3}{F}+1 }\\\ 50.8 = \frac{100}{\frac{72.3+F}{F} } \\\\50.8 = \frac{100 X F}{72.3+F}\\ \\50.8 X 72.3 + 50.8 X F = 100F\\\\3672.84 = 100F-50.8F\\3672.84 = 49.2F\\F = 74.65g$