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Marysya12 [62]
2 years ago
13

Ultraviolet light from a distant star is traveling at 3.0 × 108 m/s. how long will it take for the light to reach earth if it mu

st travel 4.0 × 1013 km? 2.7 × 10–2 hours 2.2 × 103 hours 3.7 × 104 hours 1.3 × 105 hours
Physics
1 answer:
castortr0y [4]2 years ago
5 0

Given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.

<h3>What is Speed?</h3>

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

  • Speed of the Ultraviolet light c = 3.0 × 10⁸m/s = 1.08 × 10⁹km/h
  • Distance it must cover d = 4.0 × 10¹³km
  • Time elapsed t = ?

We substitute our given values into the expression above.

Speed = Distance ÷ time

1.08 × 10⁹km/h = 4.0 × 10¹³km ÷ t

t = 4.0 × 10¹³km ÷ 1.08 × 10⁹km/h

t = 3.7 × 10⁴ hrs

Therefore, given speed and the distance that must be covered, the time it will take the ultraviolet light to reach the earth is 3.7 × 10⁴ hours.

Learn more about speed here: brainly.com/question/7359669

#SPJ4

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2 years ago
50 kg of water at 75o C is cooled to 25o C. How much heat was given off?
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b the answer is b

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2 years ago
Why are some forces considered to be noncontact forces? A. Objects must be far apart in order to exert a force. B. Objects do no
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Answer:

B. Objects do not have to touch each other to experience a force.

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3 years ago
A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway
Andru [333]

Answer:

<em>63.44 rad/s</em>

<em></em>

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet v_{1} = 250 m/s

final velocity of bullet v_{2} = 140 m/s

loss of kinetic energy of the bullet = \frac{1}{2}m(v^{2} _{1} - v^{2} _{2})

==> \frac{1}{2}*0.0033*(250^{2}  - 140^{2} ) = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = \frac{1}{2} mv^{2}

70.785 = \frac{1}{2}*0.25*v^{2}

566.28 = v^{2}

v= \sqrt{566.28} = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 =<em> 63.44 rad/s</em>

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At the beach, differences in temperatures create wind which transfers heat between the land and the water. The same process is a
NeTakaya

Heat transfer by means of fluid flow is called "convection".

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Read 2 more answers
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