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Brut [27]
2 years ago
6

An unfortunate astronaut loses his grip during a spacewalk and finds himself floating away from the space station, carrying only

a rope and a bag of tools. First he tries to throw a rope to his fellow astronaut, but the rope is too short. In a last ditch effort, the astronaut throws his bag of tools in the direction of his motion, away from the space station. The astronaut has a mass of ma=102 kgma=102 kg and the bag of tools has a mass of mb=10.0 kg.mb=10.0 kg. If the astronaut is moving away from the space station at vi=2.10 m/svi=2.10 m/s initially, what is the minimum final speed vb,fvb,f of the bag of tools with respect to the space station that will keep the astronaut from drifting away forever?
Physics
1 answer:
alexira [117]2 years ago
3 0

Answer:

The answer is "2.352 \ \frac{m}{s}"

Explanation:

\to mass(m_1)=102 \ kg\\\\\to  mass(m_2)=10 \ kg \\\\\to v=2.10\ \frac{m}{s}\\\\

momentum before:

\to p=(m_1+m_2)v

       =(102+10)2.10\\\\=(102\times 2.10 +10 \times 2.10)\\\\=214.2+21\\\\=235.2

momentum After:

\to p=(m_1+m_2)v

       =(102\times 0 +10 \times v)\\\\ =(0 +10v)\\\\=10v\\

Calculating the conservation of momentum:

\to \text{momentum before = momentum After}

\to 235.2=10v\\\\\to v= \frac{235.2}{10}\\\\ \to v=2.352 \ \frac{m}{s}

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An airplane has a takeoff speed of 62 m/s. assuming a constant acceleration of 1.7 m/s2, what is the minimum length of runway it
UkoKoshka [18]

Data:

u=0 m/s is the initial velocity of the plane

v=62 m/s is the final velocity of the plane (at which the plane takes off)

a=1.7 m/s^2 is the acceleration of the plane

To find the minimum distance S the plane needs to take off, we can use the following equation:

2aS=v^2 -u^2

Re-arranging it and substituting the numbers, we find

S=\frac{v^2-u^2}{2a}=\frac{(62 m/s)^2-0}{2(1.7 m/s^2)}=1131 m

3 0
3 years ago
A slingshot is used to shoot a small ball upward into the air; when the pouch is released, the sling propels the ball upward. as
alexdok [17]

Explanation:

kinetic to gravitational

4 0
3 years ago
Studying neutrinos helped to explain how our Sun works but led to changes in theories of particle physics, how is this process c
julsineya [31]

Explanation:

Yes,  evidently, the process is consistent with the scientific process because in scientific process falsification and modification are two very important traits. So this new concept have modified the existing theories.

Through the modification a theory is adjusted without undermining other discovering made through the theory's prediction.

There are many other evidences that prove this fact for example Einstien's Theory of relativity also changes the existing concepts.

3 0
2 years ago
The speed of sound in room temperature (20°C) air is 343 m/s; in room temperature helium, it is 1010 m/s. The fundamental freque
Lera25 [3.4K]

Answer: f = 927.55Hz

Explanation: Since the the tube is open-closed, the length of air and the wavelength of sound passing through the tube is given below

L = λ/4 where λ = wavelength.

speed of sound in air = v = 343m/s.

fundamental frequency of open closed tube = 315Hz

λ = 4L.

v = fλ

343 = 315 * 4L

343 = 1260 * L

L = 343/ 1260

L = 0.27m

In the same tube of length L = 0.27m but different medium ( helium), the speed of sound is 1010m/s.

The length of tube and wavelength are related by the formulae below

L = λ/4, λ=4L

λ = 4 * 0.27

λ = 1.087m.

v = fλ

1010 = f * 1.087

f = 1010/1.807

f = 927.55Hz

4 0
3 years ago
irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano
Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

            p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

  = 1 / (0.042)                                     [ 1 parsecs = 1 arcseconds ]

  = 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

<u>brainly.com/question/16471213</u>

#SPJ4

6 0
1 year ago
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