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taurus [48]
3 years ago
9

Light is an oscillating magnetic field alone. an oscillating electric field alone. electric and magnetic fields that oscillate p

arallel to each other. electric and magnetic fields that oscillate perpendicularly to each other. electric and magnetic fields that oscillate in a random direction relative to each other. Which of the statements concerning light are true? The speed of light is the same no matter what material it is traveling through. Its propagation direction is perpendicular to both the electric field and the magnetic field. Its propagation direction is parallel to both the electric field and the magnetic field. The speed of light in matter is less than the speed of light in a vacuum. It moves at a constant speed through a vacuum. The speed of light in matter is greater than the speed of light in a vacuum.
Physics
1 answer:
cupoosta [38]3 years ago
7 0

Explanation:

  • As light is an electromagnetic wave that causes sensation of sight.  
  • Every electromagnetic wave has electrical and magnetic components in it which are perpendicular to the direction of propagation of wave.  
  • The electrical and magnetic components are mutually perpendicular to each other in an electromagnetic wave.
  • The speed of light in vacuum is c=3\times 10^{8}\ m.s^{-1}, which decreases when it passes through a matter (precisely a transparent medium) which allows the passage of light through it. When the light passes through some transparent matter then its speed decreases as compared to the speed of light in vacuum.
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Likurg_2 [28]

Answer:

Part a)

T = 0.81 s

Part b)

v_x = 3.33 m/s

Part c)

v_y = 3.91 m/s

Part d)

\theta = 49.55 degree

Part e)

T = 1.11 s

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

\Delta y = 1.80 - 1.02

\Delta y = 0.78 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(0.78)

v_y = 3.91 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 3.91 - 9.81 t_1

t_1 = 0.39 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(1.80 - 0.95)

v_y = 4.08 m/s

time taken by it to reach this height

4.08 = v_i + at

4.08 = 0 + 9.81 t_2

t_2 = 0.42 s

T = t_1 + t_2

T = 0.81 s

Part b)

Horizontal velocity

v_x = \frac{x}{t}

v_x = \frac{2.70}{0.81}

v_x = 3.33 m/s

Part c)

vertical velocity is the intial y direction velocity

v_y = 3.91 m/s

Part d)

Take off angle is given as

tan\theta = \frac{3.91}{3.33}

\theta = 49.55 degree

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

\Delta y = 2.50 - 1.20

\Delta y = 1.30 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(1.30)

v_y = 5.05 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 5.05 - 9.81 t_1

t_1 = 0.51 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(2.50 - 0.72)

v_y = 5.9 m/s

time taken by it to reach this height

5.9 = v_i + at

5.9 = 0 + 9.81 t_2

t_2 = 0.60 s

T = t_1 + t_2

T = 1.11 s

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Answer:

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Answer:

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