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3 years ago

15

A 4 kg block slides down a rough inclined plane inclined at 30° 179009.GIF with the horizontal. Determine the coefficient of kin

etic friction between the block and the surface if the block has an acceleration of 1.2 m/s2.

1 answer:

mrs_skeptik [129]3 years ago

5 0

F=ma
F= 4x1.2
F= 4.8 N

F= 4gsin30 - Friction
Friction= 19.6 - 4.8 N
Friction= 14.8 N

Friction= u x 4gcos30
14.8 / 4gcos30 = u
u= 0.43596...
u= 0.44

coefficient is 0.44

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Calculate the mag-netic field (magnitude and direc-tion) at a point p due to a current i=12.0 a in the wire shown in fig. p28.68

creativ13 [48]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The magnitude is $B= 4.2 *10^ {-6}T$ , the direction is into the page

Explanation:

From the question we are told that

The current is $i = 12.0 A$

The radius of arc bc is $r_{bc} = 30.0 \ cm =\frac{30}{100} = 0.3m$

The radius of arc da is $r_{da} = 20.0 \ cm = \frac{20}{100} = 0.20 \ m$

The length of segment cd and ab is = $l = 10cm = \frac{10}{100} = 0.10 m$

The objective of the solution is to obtain the magnetic field

Generally magnetic due to the current flowing in the arc is mathematically represented as

$B = \frac{\mu_o I}{4 \pi r}$

Here I is the current

$\mu_o$ is the permeability of free space with a value of $4\pi *10^{-7}T \cdot m/A$

r is the distance

Considering Arc da

$B_{da} = \frac{\mu_o I}{4 \pi r_{da}} \theta$

Where $\theta$ is the angle the arc da makes with the center from the diagram its value is $\theta = 120^o = 120^o * \frac{\pi}{180} = \frac{2\pi}{3} rad$

Now substituting values into formula for magnetic field for da

$B_{da} = \frac{4\p *10^{-7} * 12}{4 \pi (0.20)}[\frac{2 \pi}{3} ]$

$= \frac{10^{-7} * 12}{0.20} * [\frac{2 \pi}{3} ]$

$B_{da}= 12.56*10^{-6} T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to da is into the pages of the paper

Considering Arc bc

$B_{bc} = \frac{\mu_o I}{4 \pi r_{bc}} \theta$

Substituting value

$B_{bc} = \frac{4 \pi *10^{-7} * 12}{4 \pi (0.30)} [\frac{2 \pi}{3} ]$

$B_{bc}= 8.37*10^{-6}T$

Looking at the diagram to obtain the direction of the current and using right hand rule then we would obtain the the direction of magnetic field due to bc is out of the pages of the paper

Since the line joining P to segment bc and da makes angle = 0°

Then the net magnetic field would be

$B = B_{da} - B{bc}$

$= 12.56*10^{-6} - 8.37*10^{-6}$

$= 4.2 *10^ {-6}T$

Since $B_{da} > B_{bc}$ then the direction of the net charge would be into the page

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