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liberstina [14]

1 year ago

15

5. A 1-kg car and a 2-kg car are both released from the top of the same hill and roll down a frictionless track. At the bottom o

f the hill, which car will have the greater
speed? Which car will have the greater Kinetic energy?

O A The 1-kg car will have greater speed, and the 2-kg car will have greater kinetic energy

OB. The 2 kg car car will have greater speed, and the 7-kg car will have greater kinetic energy

OC. The cars will have equal speeds and equal kinetic energies.

OD. The cars will have equal speeds and the 2 kg car will have greater kinetic energy

1 answer:

grigory [225]1 year ago

7 0

The cars will have equal speeds and the 2 kg car will have greater kinetic energy.

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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

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Complete Question

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

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

Explanation:

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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 = $l = 10cm = \frac{10}{100} = 0.10 m$

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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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