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Oxana [17]
3 years ago
10

A 2 kg object is subjected to three forces that give it an acceleration −→a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj. If two of the thre

e forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = −(12.0N)ˆi + (8.00N)ˆj, find the third force.
Physics
1 answer:
Angelina_Jolie [31]3 years ago
7 0

Answer:

\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}

Explanation:

You have three forces F1, F2 an F3 that produce the following  acceleration:

a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj

you know that force F1 and F2 are:

F1 = (30.0N)ˆi + (16.0N)ˆj

F2 = −(12.0N)ˆi + (8.00N)ˆj

and the force F3 is unknown:

F3 = F3x ˆi + F3y ˆj

The second Newton law is given by the following equation:

\vec{F}=m\vec{a}

F: the total force = F1 +F2 + F3

m: mass of the object = 2 kg

By the properties of vectors you have:

\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}

Both x and y component must be equal in the previous equality, then you have:

18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N

Hence, the vector F3 is:

\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}

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Answer:

change in internal energy 3.62*10^5 J kg^{-1}

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What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
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Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

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