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Oksi-84 [34.3K]
3 years ago
11

When the car was stopped by the tree, its change in velocity during the collision was -6 meters/second. This change in velocity

occurred in 2 seconds. What was the acceleration of the car?
Physics
1 answer:
jonny [76]3 years ago
8 0
6 meters per second

You will need to add six two times to find the acceleration
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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

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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