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BartSMP [9]
3 years ago
13

The stars, Rigel and Betelgeuse, are both found in the constellation Orion. Rigel is a blue supergiant, and Betelgeuse is a red

supergiant. Which of the following correctly compares the temperatures of Rigel and Betelgeuse?
Physics
1 answer:
SIZIF [17.4K]3 years ago
6 0

Answer:

batrix

Explanation:

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B I think for 19 and D for 20
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Why does air blow out of your lungs when you exhale
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When breathing out, it pushes your diaphragm, the muscle below your lungs, up, which then causes the air to leave your lungs.
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To which series would the emitted light belong if an electron in a hydrogen atom underwent a transition from level n = 5 to leve
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So we want to know in which series would the emitted light belong if the electron in a hydrogen atom would go from the energy level n=5 to n=1. The correct answer is Lyman series. Complete Lyman series is from energy level n=6 to n=1. All of the wavelengths in the Lyman series are in the ultraviolet part of the electromagnetic spectrum. 
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4 years ago
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
Arisa [49]

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

              = 1/2 (6 X 15)

              = 1/2 (90)

              = 45 meters

Hence, the object traveled 45 meters.

4 0
4 years ago
A baseball pitcher throws a baseball horizontally at a linear speed of 42.5 m/s (about 95 mi/h). Before being caught, the baseba
Xelga [282]

Answer:

v_t=46.4532\ m.s^{-1}

Explanation:

Given:

  • Linear horizontal speed of the ball, v_x=42.5\ m.s^{-1}
  • distance travelled by the ball, x=17.4\ m
  • angle rotated by the ball, \theta=44.1\ rad
  • radius of the ball, r=3.67\ cm=0.0367\ m

<u>the time for which the ball travels:</u>

t=\frac{x}{v_x}

t=\frac{17.4}{42.5}

t=0.4094\ s

<u>Angular speed of the ball:</u>

\omega=\frac{\theta}{t}

\omega=\frac{44.1}{0.4094}

\omega=107.7155\ rad.s^{-1}

<u>Now the tangential speed at the equator of the ball:</u>

v_t=r.\omega+v_x (the tangential speed due to rotation and the linear speed are having the same direction)

v_t=0.0367\times 107.7155+42.5

v_t=46.4532\ m.s^{-1}

4 0
4 years ago
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