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umka2103 [35]
2 years ago
10

A spaceship travels at approximately 0. 25c with respect to earth. this means 24 hours at the spaceship will be x hours to us on

earth. find the time passed ton earth
Physics
1 answer:
Blababa [14]2 years ago
4 0

The time passed on earth is  mathematically given as

t' = 24.79 hrs

<h3>What is the time passed on earth?</h3>

Generally, the equation for is time  mathematically given as

t' = \gamma t

Where

\gamma = Lorentz\ factor \\\\\gamma = 1/ \sqrt {(1 - v^2/c^2)}

t' = t/ \sqrt {(1 - v^2/c^2)}

Therefore

t' = 24/ \sqrt {(1 - (0.25c/c)^2) }

t'= 24/ \sqrt {(1 - 0.25^2)

t' = 24.79 hrs

In conclusion, the time passed on earth

t' = 24.79 hrs

Read more about the time

brainly.com/question/28050940

#SPJ1

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1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
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The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

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Speed of jumbo jet in southwesterly direction (v_j) = 550 mph

Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

Similarly, the velocity of the stream is, \vec{v_s}=80\vec{i}

Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

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