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3 years ago

Discuss how the move over law applies to this is situation and your responsibility as a driver

Physics

2 answers:

lesya [120]3 years ago

6 0

Answer:

There are certain rules which are required to be obeyed by any person who is on the road while driving any sort of vehicle i.e a bike, cycle or a sedan etc. Each road is divided in to three main parts the emergency line, the high speed and the low speed line.

So, as according to the mover over law if a driver at first is traveling by the emergency line and there comes any sort of government , police, any two truck, fire-brigade or an ambulance he or she must vacate that emergency line for the mentioned vehicles as, that are termed to be treated more differently just because of there work nature.

Liono4ka [1.6K]3 years ago

3 0

Answer:

The two reasons are as given.

Explanation:

The move over law is defined for safety of all the personal involved. This includes the officer, the driver and the person being pulled over. As a driver pulling over helps the officer control the situation with the pulled over person
If it is an ambulance or a firetruck with their sirens on it means they are having to get to an emergency so for the others safety as well as for the emergency to save lives pulling over is required so that the emergency vehicle can move further.

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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const

grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr. and Area of this element is

$dA=2\pi r dr$

since current density is given

$J=kr$

then , current through this element will be,

$di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr$

integrating on both sides between the appropriate limits,

$\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr
\\\\
I=\frac{2\pi\,ka^3}{3} -------------------------------(1)$

Magnetic field can be found by using Ampere's law

$\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}$

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

$\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

$B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=
\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}
\\\\B_{in}=\frac{\mu_0kl^2}{3}
$

now using equation 1, putting the value of k,

$B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}
\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}
$

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

$\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}
$

by symmetry $\vec{B}$ will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

$B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)
\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}
\\\\B_{out}=\frac{\mu_0ka^3}{3r}
$

again using,equaiton 1,

$B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}
\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}$

8 0

3 years ago

Letter C on the map labels which feature of the Middle East?

likoan [24]

Honest, the map is so tiny, and so fuzzy when I blow it up, I really can't see anything on it clearly. But I think maybe I do see a letter ' C ' in the eastern Mediterranean, with a curved line over to the southern Gaza strip, where it meets Sinai. So I'll say it's the Gaza Strip.

6 0

3 years ago

A magnetic force can act on an electron even when it A) is at rest B) moves parallel to magnetic field lines C) both of these D)

Kobotan [32]

Answer: A)

Explanation: when an electron is placed in a magnetic field, it experiences a force.

This force is given below as

F=qvB*sinθ

F = force experienced by charge.

q = magnitude of electronic charge

v = speed of electron

B= strength of magnetic field

θ = angle between magnetic field and velocity.

What defines the force exerted on the charge is the angle between the field and it velocity.

If magnetic field is parallel to velocity, then it means that θ=0° which means sin 0 = 0, which means

F = qvB * 0 = 0.

The charge being at rest has nothing to do with the angle between magnetic field strength and velocity.

3 0

3 years ago

Objects in free fall are weightless. true or false???

frez [133]

The answer is false

7 0

3 years ago

Researchers want to see if college students are more committed to their fraternity after going through a hazardous hazing ritual

Mila [183]

Answer:

College student's commitment

Explanation:

The dependent variable is the college students since it is the variable we are going to measure as a result of the independent variable which is the hazardous hazing ritual or non-hazardous hazing ritual.

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3 years ago