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Whitepunk [10]
3 years ago
15

When is the net force on an object equal to zero?

Physics
2 answers:
frez [133]3 years ago
4 0
There are all kinds of conditions that could bring that about.

1. The most common one is friction. An object is moving that it is not accelerating. That means the frictional force is equal to the force causing motion.

2. The downward gravitational force is being negated by a rope suspending an object. The upward pull by the rope cancels the downward pull of gravity.

3. The rules of the game prohibit motion. A ballplayer would be stupid to leave second base while the pitcher has the ball in his hand (a physics joke. Please ignore).

4. A normal force upwards by the earth's surface is balanced by the weight of an object laying on the ground.

5. Speaking of baseball, there is no acceleration in the horizontal direction, so the net force on the ball horizontally is 0. 

6. An elastic band held and aimed horizontally. The force of the elastic to launch a spitball vs the hand holding the elastic back waiting for the best time to let the spitball go.

That should get you going.
Rufina [12.5K]3 years ago
3 0
When ALL the forces are equal, or in general when <span>forces in the opposite direction are equal

so we can have newton example:

10 newtons X -10 newtons

netforce = 0 newtons</span>
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Answer:

Explanation:

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3 years ago
During which segments is kinetic energy decreasing?
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Answer:

m = 1.26*10²⁵ kg.

Explanation:

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First of all, we need to get the value of g.

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At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

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We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

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Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for  rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

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See attachment below.

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