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4 years ago

When trying to solve a frame problem it will typically be necessary to draw many free body diagrams. a)-True b)-False

Engineering

1 answer:

klasskru [66]4 years ago

6 0

Answer:

True

Explanation:

When trying to solve a frame problem in Engineering or Physics, it will typically be necessary to draw more than one body diagram.

When we have several parts of the frame or a set of frames, we have the anchor point, as well as the intersections of frames. Besides that, usually, there is a particle or rigid body together with the frame system. In this sense, usually, it is required to analyze a body diagram for the particle or rigid body suspended, as well as the intersections of the frames. So, usually, it will be required a minimum of two body diagrams.

If the system is more complex, or there are many intersections points, it will be required more than two body diagrams.

Finally, indeed, it will typically be necessary to draw many-body diagrams.

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netineya [11]

B. false

Explanation: because it is

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3 years ago

For each of the resistors shown below, use Ohm's law to calculate the unknown quantity, Be sure to put your answer in proper eng

daser333 [38]

Answer:

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Explanation:

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3 years ago

**Please Help. ASAP**

natima [27]

Answer:

The answer is below

Explanation:

$\frac{v-u}{a} =t\\\\Making \ v\ the \ subject\ of\ formula:\\\\First \ cross-multiply:\\\\v-u=at\\\\add\ u\ to \ both\ sides:\\\\v-u+u=at+u\\\\v=u+at$

$\frac{y-x^2}{x}=3z\\ \\Making\ y\ the\ subject\ of\ formula:\\\\First \ cross \ multiply:\\\\y-x^2=3xz\\\\y=3xz+x^2\\\\y=x(x+3z)$

$x+xy=y\\\\Making\ x\ the\ subject\ of\ formula:\\\\x(1+y)=y\\\\Divide\ through\ by\ 1+y\\\\\frac{x(1+y)}{1+y} =\frac{y}{1+y} \\\\x=\frac{y}{1+y}$

$x+y=xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ x\ from \ both\ sides:\\\\x+y-x=xy-x\\\\y=xy-x\\\\y=x(y-1)\\\\Divide\ through\ by \ y-1\\\\\frac{y}{y-1} =\frac{x(y-1)}{y-1}\\ \\x=\frac{y}{y-1}$

$x=y+xy\\\\Making\ x\ the\ subject\ of\ formula:\\\\Subtract\ xy\ from \ both\ sides:\\\\x-xy=y+xy-xy\\\\x-xy=y\\\\x(1-y)=y\\\\Divide\ through\ by \ 1-y\\\\\frac{x(1-y)}{1-y} =\frac{y}{1-y}\\ \\x=\frac{y}{1-y}$

$E=\frac{1}{2}mv^2-\frac{1}{2}mu^2\\ \\Making\ u\ the\ subject \ of\ formula:\\\\Multiply \ through\ by \ 2\\\\2E=mv^2-mu^2\\\\mu^2=mv^2-2E\\\\Divide\ through\ by\ m:\\\\u^2=\frac{mv^2-2E}{m}\\ \\Take\ square\ root\ of \ both\ sides:\\\\u=\sqrt{\frac{mv^2-2E}{m}}$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\\ \\Making\ y\ the\ subject \ of\ formula:\\\\\frac{x^2}{a^2}-1=\frac{y^2}{b^2}\\\\Multiply\ through\ by\ b^2\\\\b^2(\frac{x^2}{a^2} -1)=y^2\\\\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{b^2(\frac{x^2}{a^2} -1)}$

$ay^2=x^3\\\\Make\ y\ the\ subject\ of\ formula:\\\\Divide\ through\ by\ a:\\\\y^2=\frac{x^3}{a}\\ \\Take\ square\ root\ of\ both\ sides:\\\\y=\sqrt{\frac{x^3}{a}} \\$

4 0

3 years ago

An incandescent light bulb can be regarded as a resistor. If its power output is 100W, calculate the resistance of the light bul

stira [4]

Answer: $121\ \Omega$

$0.909\ A$

Explanation:

Given

Power $P=100\ W$

Voltage applied $V=110\ V$

Resistance of the bulb is given by

$P=\frac{V^2}{R}$

$100=\frac{110^2}{R}$

$R=\frac{12100}{100}$

$R=121\ \Omega$

Current drawn by the Power source is given by

$P=V\cdot I$

$I=\frac{P}{V}$

$I=\frac{100}{110}$

$I=0.909\ A$

8 0

4 years ago

Define factors that can change the performance of a polymer, such are additives

inna [77]

Answer:

The performance of the polymer is basically change by the various type of factors like shape, tensile strength and color.

The polymer based products are basically influenced by the environmental factors like light, acids or alkalis chemicals, salts and also heat.

The additives is one of the type of chemical polymer which basically include polymer matrix for improving the ability of processing of the polymer. It also helps to enhance the production and requirement of the polymer products in the environment.

8 0

3 years ago