Respuesta :
Answer:
the time taken by the astronaut to reach safety = 9.8 hr
Explanation:
The equation for intensity can be written as :
[tex]I = \frac{P}{A}[/tex]
where :
[tex]\frac{I}{c}= \frac{F}{A}[/tex]
Replacing that into the above previous equation; we have:
[tex]\frac{P}{Ac}=\frac{F}{A}[/tex]
[tex]F = \frac{P}{c}[/tex]
However ; the force needed to push the astronaut is as follows:
F = ma
where ;
m = mass of the astronaut and a = its acceleration
we as well say;
[tex]\frac{P}{c} = ma[/tex]
[tex]a = \frac{P}{mc}[/tex]
Replacing P with 1000 W ; m with 80 kg and [tex]3*10^{8} \ m/s[/tex] for c
Then; a = [tex]\frac{1000 \ W}{(80)(3.0*10^8)}[/tex]
a = [tex]4.2*10^{-8} \ m/s[/tex]
It is also known that the battery will run for one hour and after which the battery on the laser will run out
Then to determine the change in the position after the first hour ; we have:
[tex]\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 \ h)^2[/tex]
[tex]\Delta x_1 = \frac{1}{2}*4.2*10^{-8} \ m/s^2 ) (1.0 *3600 s)^2[/tex]
= 0.27 m
Furthermore, the final velocity of the astronaut is determined as:
[tex]v_1 = at_1[/tex]
where ;
[tex]v_1 = final \ velocity[/tex]
replacing [tex]t_1 = 1.0 \ h[/tex] and a = [tex]4.2*10^{-8} \ m/s[/tex]; Then:
[tex]v_1 = (4.2*10^8 \ m/s * 1.0 \ h * \frac{ 3600\ s}{1.0 \ h})[/tex]
[tex]v_1 = 1.51 *10^{-4} \ m/s[/tex]
Also; when he drifted 5.0 m away from the capsule; the distance is far short of the 5 m but he still have 9 hours left of oxygen . In addition to that, he acceleration is also zero and the final velocity remains the same, so:
To find the final distance traveled by the astronaut ;we have:
[tex]\Delta x_2 = d - \Delta x_1[/tex]
where;
[tex]\Delta x_2[/tex] = the final distance
d = total distance
So;
[tex]\Delta x_2 = 5 m - 0.27 m \\ \\ \Delta x_2 = 4.73 \ m[/tex]
The time taken to reach the final distance can be calculated as:
[tex]t_2 = \frac{\Delta x_2 }{v_1}[/tex]
where;
[tex]t_2[/tex] = is the time to reach the final distance
Replacing 4.73 for [tex]{\Delta x_2 }[/tex] and [tex]1.51*10^{-4}[/tex] m/s for [tex]v_1[/tex]
[tex]t_2 = \frac{4.73 \ m }{1.51*10^{-4} \ m/s}[/tex]
[tex]t_2 = 31500 \ s (\frac{1.0 \ h}{3600 \ s} )[/tex]
[tex]t_2 = 8.8 \ h[/tex]
We knew the laser was operated for 1 hour; thus the total time taken by the astronaut to reach the final distance is the sum of the time taken to reach the final distance and the operated time of the laser.
Hence ; the time taken by the astronaut to reach safety = 9.8 hr
The total time spent by the astronaut before reaching the safety is 9.8 hours.
The given parameters;
- mass of the astronaut, m = 80 kg
- final position of the astronaut, x₂ = 5 m
- power of the laser, P = 1000 W
- time of operation of the laser, t = 1.0 h
The acceleration of the astronaut from the power of the laser is calculated as follows;
[tex]F = \frac{P}{c} \\\\ma = \frac{P}{c} \\\\a = \frac{P}{mc} = \frac{1000}{80 \times 3\times 10^{8}} = 4.167 \times 10^{-8} \ m/s^2[/tex]
The displacement of the astronaut in 1 hour is calculated as follows;
1 hour = 3600 s
[tex]x_1 = v_0t + \frac{1}{2} at^2\\\\x_1 = 0 + \frac{1}{2} \times (4.167 \times 10^{-8})(3600)^2\\\\x_1 = 0.27 \ m[/tex]
The change in the position of the astronaut is calculated as follows;
[tex]\Delta x = x_2 - x_1\\\\\Delta x = 5\ m - 0.27 \ m = 4.73 \ m[/tex]
The time travel the remaining 4.73 m is calculated as follows;
The final velocity of the astronaut after completing 0.27 m;
[tex]v^2 = u^2 + 2ax_1\\\\v^2 = 0 + 2(4.167\times 10^{-8})(0.27)\\\\v= \sqrt{2.2502\times 10^{-8}} \\\\v = 1.5\times 10^{-4} \ m/s[/tex]
The time of motion of the astronaut in travelling 4.73 m;
[tex]t_2 = \frac{\Delta x}{v} \\\\t_2 = \frac{4.73}{1.5\times 10^{-4}} = 31533.33\ seconds = \frac{31533.33}{3600} = 8.8 \ hr[/tex]
The total time spent by the astronaut before reaching the safety;
total = 1 hr + 8.8 hr = 9.8 hr
Thus, the total time spent by the astronaut before reaching the safety is 9.8 hours.
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