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3 years ago

How does the mass of an object affect the outcome when the forces acting upon it are unbalanced? I need six sentences...

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

6 0

<h3><u>Mass of an object affect the outcome of unbalanced forces:</u></h3>

Newton’s second law of motion deals with the result of motion of an object when unbalanced forces are applied. The second law of motion establish the relationship between mass, acceleration and unbalanced forces of the object. The below points explains the relation between unbalanced forces acting on mass of the object. The acceleration of object would be higher when the unbalanced force is higher.

The equation for the unbalanced force is,

$Unbalanced\ force = Mass \times Acceleration.$

For Example: Two masses of 500 kg and 1 kg is applied with same unbalanced forces. The change in motion of 500 kg would be very much less than the change in motion of 1 kg.

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Answer:

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A scuba diver and her gear displace a volume of 67.0 l and have a total mass of 64.0 kg. (a) what is the buoyant force on the di

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3 years ago

An airplane of mass 1.60 ✕ 104 kg is moving at 66.0 m/s. The pilot then increases the engine's thrust to 7.70 ✕ 104 N. The resis

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(a) No, because the mechanical energy is not conserved

Explanation:

The work-energy theorem states that the work done by the engine on the airplane is equal to the gain in kinetic energy of the plane:

$W=\Delta K$ (1)

However, this theorem is only valid if there are no non-conservative forces acting on the plane. However, in this case there is air resistance acting on the plane: this means that the work-energy theorem is no longer valid, because the mechanical energy is not conserved.

Therefore, eq. (1) can be rewritten as

$W=\Delta K + E_{lost}$

which means that the work done by the engine (W) is used partially to increase the kinetic energy of the airplane ( $\Delta K$ ) and part is lost because of the air resistance ( $E_{lost}$ ).

(b) 77.8 m/s

First of all, we need to calculate the net force acting on the plane, which is equal to the difference between the thrust force and the air resistance:

$F=7.70\cdot 10^4 N - 5.00 \cdot 10^4 N=2.70\cdot 10^4 N$