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3 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]3 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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Assume the following LTI system where the input signal is an impulse train (i.e.,x(t)=∑????(t−nT0)[infinity]n=−[infinity].a)Find

Igoryamba

Answer:

See explaination

Explanation:

The Fourier transform of y(t) = x(t - to) is Y(w) = e- jwto X(w) . Therefore the magnitude spectrum of y(t) is given by

|Y(w)| = |X(w)|

The phase spectrum of y(t) is given by

<Y(w) = -wto + <X(w)

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3 years ago

La base de los tema relacionados a las ciencia de las ingeniería es?

Yanka [14]

Answer:

La ciencia y la ingeniería conciben el mundo como comprensible, con reglas que gobiernan su funcionamiento y que a través de un estudio cuidadoso y sistemático se puede evidenciar mediante patrones consistentes que permitan la oportunidad de examinar las características fundamentales que mejor describen los fenómenos.

Explanation:

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2 years ago

State the number of terms for each of the following algebraic expression 2x+1

harina [27]

Answer:

Expressions are made up of terms.

A term is a product of factors.

Coefficient is the numerical factor in the term

Before moving to terms like monomials, binomials, and polynomials, like and unlike terms are discussed.

When terms have the same algebraic factors, they are like terms.

When terms have different algebraic factors, they are unlike terms.

Explanation:

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6 0

2 years ago

Assume the work done compressing the He gas is -63 kJ and the internal energy change of the gas is 79 kJ. What is the heat loss

klemol [59]

Answer:

Heat gain of 142 kJ

Explanation:

We can see that job done by compressing the He gas is negative, it means that the sign convention we are going to use is negative for all the work done by the gas and positive for all the job done to the gas. With that being said, the first law of thermodynamics equation will help us to solve this problem.

Δ $U = Q + W$ ⇒ $Q =$ Δ $U -W$

$Q = 79 - (-63) = 142 kJ$

Therefore, the gas gained heat by an amount of 142 kJ.

3 0

2 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

$R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19} } =0.93ohm*cm$

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })$

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

$w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })$

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

$V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2} } , if n_{i}=1.5x10^{10}cm^{-3} \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15} }{(1.5x10^{10})^{2} } )=0.83V$

$w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15} } } =3.35x10^{-5} cm=0.335um$

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

$Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm$

7 0

2 years ago