All categories

3 years ago

Describe the limitations of the free body diagrams.

Physics

1 answer:

Olenka [21]3 years ago

8 0

Answer:

Free-body diagrams are defined as the diagram that represents the direction and magnitude of all forces that act on an object.

There are some limitations of the free-body diagrams, that are:

The free-body diagram is based on coordinate system that increases the complexity of the diagram.
There are lot of forces acting on an object such as friction, gravity, drag, tension, and normal force and to calculate the end result, it is important to determine the correct direction of all the forces otherwise wrong direction of any one of the force can give the wrong answer.
Free-body diagrams do not depend on the size and shape of the body, that is why unable to calculate the rotation and torque.

You might be interested in

You have a 160-Ω resistor and a 0.430-H inductor. Suppose you take the resistor and inductor and make a series circuit with a vo

S_A_V [24]

Answer:

A. Z = 185.87Ω

B. I = 0.16A

C. V = 1mV

D. VL = 68.8V

E. Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

$Z=\sqrt{R^2+\omega^2L^2}$ (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

$Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega$

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

$I=\frac{V}{Z}$ (2)

V: voltage amplitude = 30.0V

$I=\frac{30.0V}{185.87\Omega}=0.16A$

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

$V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV$ (3)

D. The voltage across the inductor is:

$V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V$

E. The phase difference is given by:

$\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°$

5 0

3 years ago

The natural resistance of any object to change its speed is called _____.

Stella [2.4K]

This natural resistance is known as inertia. =)

6 0

3 years ago

What is the voltage, V2, in units of Volts, across resistor R2 in the circuit shown below where VS = 4V, R1 = 14 Ohms and R2 = 3

svetlana [45]

<h2>Correct answer:</h2>

$\boxed{v_{out}=2,85V}$

<h2>Explanation:</h2>

We can use voltage divider to solve this problem that is defined as the passive linear circuit producing an output voltage $v_{out}$ that is a fraction of its input voltage $v_{in}$ . So we can use the formula:

$v_{out}=\frac{R_{2}}{R_{1}+R_{2}}v_{in}, \ where \ v_{in}=v_{s}=4V \\ \\ \therefore v_{out}=\frac{35}{14+35}(4) \\ \\ \therefore v_{out}=2,85V$

3 0

3 years ago

How do i do this??????????

Sliva [168]

Power = I^2 x R
Energy = Power x Time

4 0

3 years ago

The periodic table is an arrangment of the chemical elements, ordered by atomic number, into families, and by doing so illustrat

olya-2409 [2.1K]

Answer:

<em><u>Option B) reactive to nonreactive</u></em>

Explanation:

<u>A) Solid to gas</u>

This might apply, because most of the gases are to the right portion of the periodic table

There are just a few elements that are gases at room temperature; a total of 11: six noble gases (group 18), two halogens (group 17), oxygen (group 16), nitrogen (group 15), and hydrogen (group 1).

Hydrogen is an exception to this trend because they are in the left part of the periodic table.

<u>B) Reactive to nonreactive</u>

The most reactive metals are in group 1 (alkali metals) and the most reactive nonmentals are in the group 18 (halogens). Hence, it is false that the from left to right the preperties go from reactive to nonreactive.

Hence, this is the one that cannot apply.

<u>C) Metal to metalloid to nonmetal</u>

It is true that the metallic character of the elements decreases from left to right; the metals are to the left, going to the right you find some metalloids, and further to the right you find the nonmetals.

Hence, this trend is correct.

<u />

<u>D) Increasing number of outer-shell electrons</u>

The number of outer-shell electrons definetly increase from left to right.

The number of electrons in neutral atoms equals the number of protons (to balance the positive and negative charges). Since, the atomic number (the number of protons) increase from left to right, the number of electrons also increase.

In a period (a row of the periodic table) the electrons add to the same outer-shell, then you will find that the number of electrons of the outer-shell increase from left to right. Hence, this is also a correct trend.

In conclusion, the only trend that does not apply is the described by the option B).

4 0

3 years ago

Read 2 more answers