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Tanzania [10]
3 years ago
14

when you are wearing a seat belt and need to stop suddenly, your body tries to move forward as the car slows down. you feel the

seat belt back on your body as you move forward into it. which of newtons motion laws BEST explains WHY this happens
Physics
2 answers:
REY [17]3 years ago
7 0

Answer: First Law

Explanation:

When you are sitting in the moving car, you are ina moving system, but when the car stops,  your body (that is inside the car) does not stop with it, this is because the first law, that says that an object will remain still or in constant motion until a force is applied over it.

This is why your body will continue with constant velocity after the car stops, and the seat belt keeps you in place as if you are another piece of the car that also stops with it.

chubhunter [2.5K]3 years ago
6 0

i think it would be newton's first law of motion

Newton's First Law states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

<u><em>please mark as brainliest</em></u> :) hope i helped

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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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4 years ago
Convert 1 x 10-3 nm to m. given: 1 m = 1,000,000,000 nm
vampirchik [111]

Answer:

l m = 1000000000 nm

? = 0.001

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What will a spring scale read for the weight of a 75.0-kg woman in an elevator that moves upward with constant speed of 5.8 m/s
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4 years ago
A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc
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Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

f_{c_1 = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

f_{c_1 = 3 × 10¹⁰ / 2( 0.0711  )

f_{c_1 = 3 × 10¹⁰ cm/s / 0.1422  cm

f_{c_1 =  21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( (f_{c_1 / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀e^{-\alpha _1l_{min = -100dB

we substitute

20log₁₀e^{-(310.7np/m)l_{min = -100dB

l_{min = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0
3 years ago
9- Under what circumstances would a vector have components that are equal in
valkas [14]

Explanation:

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