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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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3 a There is a thin layer of water between the blade and the ice. Suggest how this affects friction .​
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Answer:

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8 0
3 years ago
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
Please help on this one?
bezimeni [28]

Using the given equation you get:

E = 1.99x10^-25 / 9.0x10^-6

Divide 1.99 by 9.0: 1.99/9.0 = 0.22

For the scientific notation, when dividing subtract the two exponents:

25 -6 = 19

So you now have 0.22 x 10^-19

Now you need to change the 0.22 to be in scientific notation form:

2.2 x 10^-20

The answer is B.

3 0
2 years ago
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Lera25 [3.4K]
Hello There!

Your answer is Gold because it has the lowest SH

Hope This Helps You!
Good Luck :) 

- Hannah ❤
3 0
3 years ago
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