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GuDViN [60]
2 years ago
14

You have been hired to check the technical correctness of an upcoming made-for-TV murder mystery that takes place in a space shu

ttle. In one scene, an astronaut's safety line is cut while on a spacewalk. The astronaut, who is 200 meters from the shuttle and not moving with respect to it, finds that the suit's thruster pack has also been damaged and no longer works and that only 4 minutes of air remains. To get back to the shuttle, the astronaut unstraps a 10-kg tool kit and throws it away with a speed of 8 m/s. In the script, the astronaut, who has a mass of 80 kg without the toolkit, survives, but is this correct?
Physics
1 answer:
AlladinOne [14]2 years ago
5 0

Answer:

The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air

Explanation:

Due the astronaut throws the 10-kg tool kit away with a speed of 8 m/s, it gives a momentum equivalent but in the other direction, so I=mv=(10Kg)(8m/s)=80kg*m/s, then we can find the speed that the astronaut reaches due to its weight we get, v=\frac{I}{m} =\frac{80kg*m/s}{80Kg} =1m/s.

Finally, as the distance to the space shuttle is 200m, the time taken to the astronaut to reach it at the given speed will be t=\frac{d}{v}=\frac{200m}{1m/s}=200s, as the remaining air time is 4 min or 240 seconds, The astronaut who has a mass of 80 kg without the toolkit do survive with 40 seconds of remaining air.

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VikaD [51]
W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j 
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
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6 0
2 years ago
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A single slit, which is 0.050 mm wide, is illuminated by light of 550 nm wavelength. What is the angular separation between the
likoan [24]

Answer:

The separation between the first two minima on either side is 0.63 degrees.

Explanation:

A diffraction experiment consists on passing monochromatic light trough a small single slit, at some distance a light diffraction pattern is projected on a screen. The diffraction pattern consists on intercalated dark and bright fringes that are symmetric respect the center of the screen, the angular positions of the dark fringes θn can be find using the equation:

a\sin \theta_n=n\lambda

with a the width of the slit, n the number of the minimum and λ the wavelength of the incident light. We should find the position of the n=1 and n=2 minima above the central maximum because symmetry the angular positions of n=-1 and n=-2 that are the angular position of the minima below the central maximum, then:

for the first minimum

a\sin \theta_1=(1)\lambda

solving for θ1:

\theta_1=\arcsin (\frac{\lambda}{a})=\arcsin (\frac{550\times10^{-9}}{0.05\times10^{-3}})

\theta_1=0.63 degrees

for the second minimum:

a\sin \theta_2=(2)\lambda

\theta_2=\arcsin (\frac{2\lambda}{a})=\arcsin (\frac{2*550\times10^{-9}}{0.05\times10^{-3}})

\theta_2=1.26 degrees

So, the angular separation between them is the rest:

\Delta \theta =1.26-0.63

\Delta \theta=0.63

4 0
2 years ago
A plane increases its speed from 450 km/hr to 750 km/hr in 0.30 hour. What is its acceleration?
ryzh [129]

Answer:

The acceleration is found as:

a = 1000 km/h²

Explanation:

Initial speed of the plane = 450 km/h

Final speed of the plane = 750 km/h

Time taken = 0.3 h

Acceleration can be defined as the change of speed of the object divided by the time taken to bring that change

Acceleration is given as:

a=\frac{v_f-v_i}{t}

where v(f) = 750 km/h , v(i) = 450 km/h , t = 0.3 h

Substitute the values in the equation of acceleration

a= \frac{750-450}{0.3}\\a=\frac{300}{0.3}\\a=1000 km/h^2

This is the found result

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2 years ago
Dave and Alex push on opposite ends of a car that has a mass of 875 kg. Dave pushes the car to the right with a force of 250 N,
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B but im not so suree
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2 years ago
A thin, uniform rod is hinged at its midpoint. To begin with, one-half of the rod is bent upward and is perpendicular to the oth
Dima020 [189]

Answer:

\omega_1 = 6.06 rad/s

Explanation:

Initially rod is bent into L shaped at mid point

So we will have rotational inertia of the rod given as

I = \frac{mL^2}{3} + (\frac{mL^2}{12} + m(L^2 + \frac{L^2}{4}))

I = \frac{5mL^2}{3}

Now when rod becomes straight during the rotation

then we will have

I_2= \frac{(2m)(2L)^2}{3} = \frac{8mL^2}{3}

now from angular momentum conservation we have

I \omega_o = I_1 \omega_1

(\frac{5mL^2}{3}) (9.7) = \frac{8mL^2}{3}\omega_1

\omega_1 = 6.06 rad/s

7 0
2 years ago
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