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irina [24]
3 years ago
11

Time dilation means that

Physics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

e) True. when the observer is in movement the time passes slower

Explanation:

Newton's mechanics clearly established that the laws of physics are the same in all inertial frames of reference, that is, they do not have acceleration.

In subsequent experiments it was discovered that the speed of light is a universal constant and does not depend on the relative speed of the reference system where it is being measured.

Special relativity takes these two principles and uses them to analyze how a phenomenon is seen from frames of reference with relative speed between them, since the information is propagated with a finite speed (finite time).

Based on this it follows that when an observer who does not move with respect to the phenomenon measures a time called proper time (τ) and when the observer is in motion with respect to the phenomenon, the change time was due to the time it takes for the light to carry the information, for this observer with relative speed the time is

    t = τ / √(1 - (v/c)²)

where you can see that always t> τ,

let's analyze the statements given

a) False. If the clocks are at rest, they both measure their own time

b) False. Time changes due to the relative speed of the reference frames

c) False. It is a real fact because the information takes time to reach the observer

d) False. It's the opposite

e) True. It is in agreement with the superior equation of change of the time, when the observer is in movement the time passes slower

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P=\dfrac{800}{\eta}\\\Rightarrow P=\dfrac{800}{0.34}\\\Rightarrow P=2352.94117\ MW

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The number of homes that could be heated with the waste heat of this one power plant is 77647

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A) Determine the x and y-components of the ball's velocity at t = 0.0s, 2.0, 3.0 secs.
malfutka [58]

The kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7º

given parameters

  • the initial velocity of the body vₓ = 1.8m / s and v_y = 1.2 m / s
  • the movement times t = 1.0s, 2.0s and 3.0 s

to find

    a) position

    b) acceleration

    c) launch angle

Projectile launch is an application of kinematics to the movement of the body in two dimensions where there is no acceleration on the x axis and the y axis has the planet's gravity acceleration

b) To calculate the acceleration of the plant acting on the y-axis, we use that the vertical velocity of the body at the highest point is zero.

         v_y = v_{oy} - g t

where v and v({oy}  are the velocities of the body, g the acceleration of the planet's gravity and t the time

          0 = v_{oy} - gt

           g = v_{oy} / t

from the graph we observe that the highest point occurs for t = 2.0 s

           g = 1.2 / 2.0

           g = 0.6 m / s²

 

a) The position is requested for several times

X axis

in this axis there is no acceleration so we can use the uniform motion relationships

          vₓ = x / t

          x = vₓ t

where x is the position, vx is the velocity and t is the time

we calculate for the time

t = 0.0 s

          x₀ = 0

           

t = 2.0 s

          x₂ = 1.8 2

          x₂ = 3.6 m

t = 3.0 s

          x₃ = 1.8 3

          x₃ = 5.4 m

Y axis

In this axis there is the acceleration of the planet, let us use for the position the relation

          y = v_{oy} t - ½ g t²

t = 0.0 s

          y₀ = 0

          y₀ = 0 m

t = 2.0 s

         y₂ = 1.2 2 - ½ 0.6 2²

         y₂ = 1.2 m

t = 3.0 s

        y₃ = 1.2  3 - ½  0.6  3²

        y₃ = 0.9 m

c) the launch angle use the trigonometry relation

        tan θ = \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{v_y}{v_x}

        θ = tan⁻¹ \frac{1.2}{1.8}

        θ = 33.7º

measured counterclockwise from the positive side of the x-axis

With the kinematic relationships we can find the position, acceleration and launch angle of the body on the planet Exidor.

a) the position are

      time (s)  x (m)   y(m)

        0            0          0

        2.0         3.6        1.2

        3.0         5.4        0.9

b) The aceleration is  g = 0.6 m / s²

c) The launch angle      θ = 33.7ºto)

learn more about projectile launch here:

brainly.com/question/10903823

4 0
3 years ago
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