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Troyanec [42]
2 years ago
7

The fictional rocket ship Adventure is measured to be 50 m long by the ship's captain inside the rocket.When the rocket moves pa

st a space dock at 0.5c, space-dock personnel measure the rocket ship to be 43.3 m long. Part A The rocket ship Adventure travels to a star many light-years away, then turns around and returns at the same speed. When it returns to the space dock, who would have aged less: the space-dock personnel or ship's captain?
Physics
1 answer:
tensa zangetsu [6.8K]2 years ago
3 0

Complete question:

Part A:) The fictional rocket ship Adventure is measured to be 50 m long by the ship's captain inside the rocket.When the rocket moves past a space dock at 0.5c , space-dock personnel measure the rocket ship to be 43.3 m long. The rocket ship Adventure travels to a star many light-years away, then turns around and returns at the same speed. When it returns to the space dock, who would have aged less: the space-dock personnel or ship's captain?

Part B: What is the momentum of a proton traveling at 0.62 c ?

Answer

a)Who would have aged less=The Captain would have aged less

b) p=3.96*10^{-19}kgm/s

Explanation:

From the question we are told that

Length measured by captain l_c=50m

Speed of rocket past tje space dock V=0.5c

Length measured by space-dock personnel l_c=43.3m

a)

Generally time moves slower when moving at speed of light, due to time dilation or variation.

Who would have aged less=The Captain would have aged less

b)

Generally the equation for Relativistic Momentum  is mathematically given as

p=\frac{m*v}{1 - v^2/c^2}

p=\frac{1.67*10^2-27*0.62*3.0*10^8)}{\sqrt{ 1 -0.6^2}}

p=3.96*10^{-19}kgm/s

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Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest. Which best describes h
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8.1km/h Northwest

Explanation:

The 8.1km/h Northwest gives the best description of her distance from start to finish. This distance can be represented in a right angle triangle , this is the hypotenuse which is the longest side of the triangle. If we add 5.7 and 5.8 this gives 11.5km/h compared to 8.1km/h which is a smaller distance and the best.

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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
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Before the engines fail, the rocket's altitude at time <em>t</em> is given by

y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

and its velocity is

v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t

The rocket then reaches an altitude of 1150 m at time <em>t</em> such that

1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2

Solve for <em>t</em> to find this time to be

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At this time, the rocket attains a velocity of

v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}

When it's in freefall, the rocket's altitude is given by

y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity, and its velocity is

v_2(t)=124\dfrac{\rm m}{\rm s}-gt

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for y_2(t) to reach 0:

1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

{v_f}^2-{v_i}^2=2a\Delta y

where v_f and v_i denote final and initial velocities, respecitively, a denotes acceleration, and \Delta y the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means y_2 will contain the information we need to find the maximum height.

-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)

Solve for y_{\rm max} and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to y_2(t)) to be about 32.6 s. Plug this into v_2(t) to find the velocity before it crashes:

v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

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Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

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Given that,

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We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

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So, the magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N. Hence, this is the required solution.

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3 years ago
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