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Dafna11 [192]
2 years ago
15

Which of the following occurs during the evolution of a star?

Physics
1 answer:
UNO [17]2 years ago
5 0
B
Hope this helps! ^-^
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Which of the following describes when light bounces off a surface
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Car headlights have both low beams and high beams. The high beams may be necessary when it is very dark outside. If the power fo
blsea [12.9K]

Answer:

Voltage, V = 12 V.

Explanation :

It is given that,

The power for the high beam is, P = 60.0 watts

Current flowing, I = 5 A

Car headlights have both low beams and high beams. The high beams may be necessary when it is very dark outside.

So, the voltage required for the high beam is 12 V.

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3 years ago
By what order of magnitude is something that runs in nanoseconds faster than something that runs in milliseconds?
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To solve this problem we will define the order of magnitude of both points, then we will obtain the radius and obtain the conclusion of the order of magnitude.

A nanosecond is one billionth of a second while and a millisecond is one millionth of a second

\frac{\text{millisec}}{\text{nanosec}} = \frac{10^{-3}}{10^{-9}} = 10^6

Therefore something that runs in nanoseconds is six times faster than something that runs in milliseconds

5 0
3 years ago
Help me with the following problem
timofeeve [1]

The rope will remain taut until the particle makes 79⁰ angle.

<h3>Change in kinetic energy of the particle</h3>

The change in kinetic energy of the particle is calculated as follows;

ΔK.E = K.Ei - K.Ef

Before the particle will achieve the given angular displacement, it will touch two new corners. Total kinetic energy lost = 30%

ΔK.E = 100%K.E - 30%K.E = 70%K.E = 0.7K.E

  • let the vertical displacement of the particle = h
  • horizontal length = side of the prism = a
  • hypotenuse side  = length of the pendulum = L
<h3>Apply principle of conservation of energy</h3>

K.E = P.E

0.7K.E = mgh

0.7(¹/₂mv²) = mg(Lsinθ)

0.7(v²) = 2g(Lsinθ)

from third kinematic equation;

v² = u² + 2gh

v² = 0 + 2gh

v² = 2g(a tanθ)

0.7(2g(a tanθ)) = 2g(Lsinθ)

0.7(a tanθ) = Lsinθ

0.7a/L = sinθ/tanθ

0.7a/L = cosθ

(0.7 x 0.8)/(3) = cosθ

0.1867 = cosθ

θ = cos⁻¹(0.1867)

θ = 79⁰

Thus, the rope will remain taut until the particle makes 79⁰ angle.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

5 0
2 years ago
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