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igomit [66]
3 years ago
8

An astronaut in deep space is at rest relative to a nearby space station. The astronaut needs to return to the space station. A

student makes the following claim: “The astronaut should position her feet pointing away from the space station. Then, she should repeatedly move her feet in the opposite direction to each other. This action will propel the astronaut toward the space station.” Is the student’s claim correct? Justify your selection.
A.) Yes. The astronaut’s feet exert a force away from the space station, creating an equal and opposite force that will accelerate the astronaut toward the space station.

B.) Yes. The astronaut’s feet will have a velocity that is transferred to her center of mass, accelerating the astronaut toward the space station.

C.) No. The astronaut’s feet are not exerting a force on another object, so there is no external force to accelerate the astronaut toward the space station.

D.) No. The astronaut would move away from the space station, not toward it, since her feet are pointed away from the space station.
Physics
2 answers:
Paha777 [63]3 years ago
7 0

Answer:

what u

Explanation:

know about rolling down in the deep when ue brain goes du m u can call that mental freeze when people talk to much puy that shi t on slow motion

Marta_Voda [28]3 years ago
4 0
Its A ig , just my guess
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stich3 [128]

The object will sail away in a straight line ... continuing in the same direction it was going when the centripetal force stopped.

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El empuje que sufre un cuerpo en un liquido es igual al pseo de volumen desalojado
lesantik [10]
Si es verdad no entiendo ni madres
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3 years ago
During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.
lord [1]

Answer:

(a) F_{sm} = 4.327\times 10^{20}\ N

(b) F_{em} = 1.983\times 10^{20}\ N

(c) F_{se} = 3.521\times 10^{20}\ N

Solution:

As per the question:

Mass of Earth, M_{e} = 5.972\times 10^{24}\ kg

Mass of Moon, M_{m} = 7.34\times 10^{22}\ kg

Mass of Sun, M_{s} = 1.989\times 10^{30}\ kg

Distance between the earth and the moon, R_{em} = 3.84\times 10^{8}\ m

Distance between the earth and the sun, R_{es} = 1.5\times 10^{11}\ m

Distance between the sun and the moon, R_{sm} =  1.5\times 10^{11}\ m

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

F_{G} = \frac{Gmm'_{2}}{r^{2}}                             (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}

F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}

F_{sm} = 4.327\times 10^{20}\ N

(b) The force exerted by the Earth on the Moon is given by eqn (1):

F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}

F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}

F_{em} = 1.983\times 10^{20}\ N

(c) The force exerted by the Sun on the Earth is given by eqn (1):

F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}

F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}

F_{se} = 3.521\times 10^{20}\ N

7 0
3 years ago
Two skaters stand facing each other. One skater's mass is 60 kg, and the other's
jeyben [28]

Answer:

v' = 2.4 m/s

Explanation:

Given that,

Mass of one skater, m = 60 kg

Mass of the other's skater, m' = 60 kg

The two skaters push off each other. After the push, the smaller  skater has a velocity of 3.0 m/s.

When there is no external force acting on a system, the momentum remains conserved. It means initial momentum is equal to the final momentum. Let v' is the velocity of the larger skater.

mv = m'v'

v'=\dfrac{mv}{m'}\\\\v'=\dfrac{60\times 3}{75}\\\\v'=2.4\ m/s

So, the velocity of the larger skater is 2.4 m/s.

3 0
3 years ago
A very long wire carries a uniform linear charge density of 5 nC/m. What is the electric field strength 13 m from the center of
kipiarov [429]

Answer:

E=6.91 N/C

Explanation:

Given that

Linear Charge density ,λ = 5 nC/m

Distance ,R= 13 m

We know that formula for long wire to find electric field

E=\dfrac{\lambda }{2\pi \varepsilon _0R}

E=Electric field

R=Distance

εo=8.85 x 10⁻¹² C²/N.m²

λ=Linear Charge density

Now by putting the values

E=\dfrac{5\times 10^{-9}}{{2\times \pi \times 8.85\times 10^{-12}\times 13}}

E=6.91 N/C

Therefore the electric filed at distance 13 m will be 6.91 N/C

5 0
4 years ago
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