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4 years ago

9

Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is

the coefficient of kinetic friction? A0.01 B0.20 C0.38 D0.48

2 answers:

k0ka [10]4 years ago

5 0

The formula is Force = coefficient of friction times the weight or normal force (in newtons). Plug the numbers in the formula. Do the simple division math. The coefficient of friction should be any number between 0 and 1. If your answer is smaller than 0 or greater than 1, go back and check your work. The answer for this question is C. 0.38

baherus [9]4 years ago

3 0

Answer:

$\mu_k = 0.38$

Explanation:

As we know that the normal force on the book is given as

$F_n = 51.7 N$

also the force required to push the book on the table is given as

$F_f = 19.5 N$

now by the formula of friction force we know that

$F_f = \mu_k F_n$

now plug in all values in it

$19.5 = \mu_k (51.7)$

here we will have

$\mu_k = \frac{19.5}{51.7}$

$\mu_k = 0.38$

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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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Stolb23 [73]

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3 years ago

The distance between a speaker and a listener is 8.0 m. Determine the frequency of the sound wave if exactly 20 waves are formed

dmitriy555 [2]

The frequency of the wave is 6800 Hz

<u>Explanation:</u>

Given:

Wave number, n = 20

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Frequency, f = ?

we know:

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3 years ago

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Crank

Answer:

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