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tiny-mole [99]
3 years ago
8

according to Newton's third law what is the equal and opposite force to the downward force of gravity pulling on a man standing

in the woods
Physics
2 answers:
Fudgin [204]3 years ago
7 0

When you're talking about gravity, it's easy to identify the equal
opposite forces.

Gravity ALWAYS produces an equal pair of opposite forces. 
They both act between the centers of the two objects, one in
each direction.

Consider the equal pair of opposite gravitational forces between
you and the Earth.  One force acts on you, and draws you toward
the center of the Earth.  We call that force "your weight". 
The other one acts on the Earth, and draws it toward the center
of you.  Hardly anybody ever talks about that one, but the two
forces are equal ... your weight on Earth is equal to the Earth's
weight on you !

RideAnS [48]3 years ago
3 0

Answer: it is the pulling upward exerted by the man on the Earth.

Explanation:

Newton's third law states that the forces come in pair: the action force and the reaction force.

Whenever an object (call it object 1) exerts a force (action force) on a second object (object 2), an equal in magnitude and opposite in direction force (reaction force) is exerted by the second object on the object 1.

Consider the force of gravity, which is the downward pulling exerted by Earth on the man standing in the woods, the action force, then the reaction force, is the counteracting force exerted by the man on the Earth, and it is a pulling of the same magnitude of the gravity pulling but upward. The downward pulling is exerted on the man, but the upward pulling is exerted on the Earth.

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Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

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Answer:

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Explanation:

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F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

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