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2 years ago

Which term is defined as the force applied to the object divided by the mass of the object?

Physics

1 answer:

Leokris [45]2 years ago

7 0

Answer:

acceleration

Explanation:

From the question: "FORCE... divided by the MASS..." is translated to a math equation as "F / m", where "F" is the force and "m" is the mass.

Now, the second Newton's law says "F = m . a ", where a is acceleration.

If you move m to the left side, clearly you have "F / m", and "a" is alone to the right. Math equation:

a = F / m

acceleration = force / mass

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

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Use Eq. cosϕ=R/Z to show that the average power delivered by the source in an L−R−C series circuit is given by Pav = I^2rmsR .

Evgen [1.6K]

Answer:

Explanation:

In a L C R circuit, the average power is given by

$P_{av}=V_{rms}I_{rms}Cos\phi$

As given in the question

CosФ = R / Z

And we know that

$V_{rms}=I_{rms}\times Z$

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Answer:

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Explanation:

check the attached file below for answer and explanation.

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