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3 years ago

11

A steady 45 N horizontal force is applied to a 15 kg object a table. The object slides against a friction force of 30 N. Calcula

te the acceleration of the object in m/s.

1 answer:

lisov135 [29]3 years ago

6 0

The net force on an object subject to friction is equal to the sum of the applied force and the frictional force.

Mathematically,

$F_{N} = ma = F_{applied} - f_{fr}$

Here, m is mass of object and a is its acceleration. We take frictional force negative because it opposes the motion of object.

Given, $m=15\ kg$ , $F_{applied} =45\ N$ and $f_{fr} = 30\ N$

Substituting these values in above formula, we get

$15\ kg\times a = 45\ N -30\ N=15\ N \\\\a=\frac{15\ N}{15\ kg} =1\ m/s^2$ .

Thus, the acceleration of an object is $1\ m/s^2.$

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Answer:

a

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b

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Explanation:

From the question we are told that

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Generally the radius is mathematically represented as

$r = \frac{d}{2}$

$r = \frac{0.03}{2}$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

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=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^{-4 } \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt }$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

So

$\mu_o I + \epsilon_o \mu _o\frac{ d \phi }{dt } = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12 } \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

So

$\mu_o I + \epsilon_o \mu _o \frac{ d [EA] }{dt } = 0$

=> $\frac{dE}{dt} =- \frac{I}{\epsilon_o * A }$

=> $\frac{dE}{dt} =- \frac{17}{8.85*10^{-12} * 7.07*10^{-4} }$

=> $\frac{dE}{dt} =- 2.72 *10^{15} \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

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Answer:

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