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11Alexandr11 [23.1K]
3 years ago
11

A space-walking astronaut has become detached from her spaceship. She's floating in space with her handy tool belt attached to h

er waist, thinking about how she might get back to the ship, which she can see 50 meters from her current location.
How can she get back to the ship? Use language from the Laws of Motion in your answer.
Physics
1 answer:
ziro4ka [17]3 years ago
8 0

Astronauts on a spacewalk use the Newton's third law for navigation. Newton's third law states that every action has an equal and opposite reaction. Action and reaction forces act on different bodies.If  a person applies a force on an object, the object applies an equal and opposite force on the person.

The astronaut who has been detached from her spaceship, has a handy tool belt attached to her waist. If she throws a tool in a direction opposite to the direction of the location of the spaceship, the tool exerts an equal force on the astronaut, propelling her in the direction of the ship. To reach the ship, she can keep throwing tools, taking care to throw it  exactly in the opposite direction in which she needs to travel.

Every time a tool is thrown backwards, the astronaut will experience a force in the forward direction in accordance with Newton's third law, ultimately enabling her to get back to the ship.

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Solar system is a part of Milkyway Galaxy.

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Two equal positive charges are held in place at a fixed distance. If you put a third positive charge midway between these two ch
eimsori [14]

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False

Explanation:

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After driving a portion of the route, the taptap is fully loaded with a total of 27 people including the driver, with an average
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We will define the Total mass to calculate the force, so our values are:M_p = 69*27=1823Kg\\M_g=15*3=45Kg\\M_c=3*5=15Kg\\M_B=25Kg

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F=mg=1948*98=19090.4N

Through the hook's Law we calculate X.

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Three charges lie along the x axis. The positive charge q1 = 15 μC is at x= 2.0 m and the positive charge q2 = 6.0μC is at the o
Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.

We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

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x₂ = -3.44m

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x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

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