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3 years ago

ASAP please!! The force vectors on an aircraft are as shown. Find the net (resultant force).

Physics

1 answer:

ioda3 years ago

6 0

Answer:

The magnitude of the net force is 5430N

Explanation:

I suggest to define the axes as aligned to the axis of the plane. This will require you to decompose only one vector, namely the Weight. We need two components of the W force: one in horizontal direction of the plane, the other perpendicular to it. Through a simple triangle argument you will se that the plane-horizontal component of W is

$W_D=3600 N\cdot\sin 27^\circ=1634N$

acting in the direction of the Drag, and the plane-perpendicular component is:

$W_L=-3600N\cdot\cos 27^\circ=-3208N$

with negative sign since it counteracts the Lift.

So the components of the netforce F are:

$F_h=T-D-W_D=(8000-1000-1634)N=5366N\\F_v=L+W_L=(4100-3208)N=829N$

The magnitude of the net force is:

$|F|=\sqrt{5366^2+829^2}N = 5430N$

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The complete question is as follows:

A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s.

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The correct answer is 0.034 J.

Explanation:

Given :

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When the position of the toy from the equilibrium is x = 0.012m, then the

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Hence by substituting the numerical values in equation 1, we get

$E= \frac{1}{2} (0.15* 0.4) + \frac{1}{2} (300* 0.012)$

= 0.034 J

Thus, the correct answer is 0.034 J.

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