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MaRussiya [10]
3 years ago
8

g An astronaut must journey to a distant planet, which is 189 light-years from Earth. What speed will be necessary if the astron

aut wishes to age only 12 years during the round trip? (Give your answer accurate to four decimal places.)
Physics
1 answer:
Butoxors [25]3 years ago
5 0

Answer:

The value is v  =  2.999 *10^{8} \  m/s

Explanation:

From the question we are told that

   The time taken to travel to the planet from earth is t = 189 \ light-years

    The  time to be spent on the ship is  t_{s} =  12 \  years

Generally speed can be obtained using the mathematical relation represented below

       t_s  =  2 * t *  \sqrt{1 -  \frac{v^2}{c^2 } }

The 2 in the equation show that the trip is a round trip i.e going and coming back

=>    12 =  2 * 189 *  \sqrt{1 -  \frac{v^2}{(3.0*10^{8})^2 } }

=>     v  =  2.999 *10^{8} \  m/s

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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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Answer:

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:

Given:

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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