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Gelneren [198K]
3 years ago
5

Why are the action and reaction forces described by Newton’s third law of motion excluded in the free-body diagram of an object?

Physics
2 answers:
Lorico [155]3 years ago
6 0

The correct answer is:

B. They act on different objects, so they would not appear together.

Let's take for example a box sitting on a table. If we draw the free-body diagram of the box, we would draw two forces:

- the weight of the box (downward)

- the normal reaction of the table on the box (upward)

However, this is not a pair of action-reaction forces as described in Newton's third law. In fact, the correct action-reaction pair would be:

- weight of the box (=gravitational force exerted by the Earth on the box)

- gravitational force exerted by the box on the Earth

We see that these two forces act on two different objects (the first one on the box, the second one on the Earth), so they do not appear in the same free-body diagram.

umka2103 [35]3 years ago
4 0
B. <span>They act on different objects, so they would not appear together.</span>
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An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed incre
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a.     F = 2.32*10^-18 N

b.     The force F is 2.59*10^11 times the weight of the electron

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a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:

v^2=v_o^2+2ax         (1)

v: final speed of the electron = 6.60*10^5 m/s

vo: initial speed of the electron = 4.00*10^5 m/s

a: acceleration of the electron = ?

x: distance traveled by the electron = 5.40cm = 0.054m

you solve the equation (2) for a and replace the values of the parameters:

a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}

Next, you use the second Newton law to calculate the force:

F=ma

m: mass of the electron = 9.11*10^-31kg

F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N

The magnitude of the force exerted on the electron is 2.32*10^-18 N

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The quotient between the weight of the electron and the force F is:

\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}

The force F is 2.59*10^11 times the weight of the electron

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