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2 years ago

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

Physics

2 answers:

lesya [120]2 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]2 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 charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

6 0

2 years ago

A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the

bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = $2\times 10^{-3}$

Extended by = 0.25mm = $0.25\times 10^{-3}$

Based on the above information, the calculation is as follows

a. The Stress of the wire is

$= \frac{force\ applied}{area\ of \ circle}$

here area of circle = perpendicular to the are i.e cross-sectional i.e

= $\frac{\pi d^{2}}{4}$

= $\frac{\pi(2\times 10^{-3})^2}{4}$

Now place these above values to the above formula

$= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}$

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

$= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}$

= $5 \times 10^{-5}$

3 0

2 years ago

A 7.00-kg object accelerates from rest to a final velocity in 55 seconds. If the magnitude of the

Len [333]

Answer:

The final velocity of the object is 330 m/s.

Explanation:

To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:

F = ma

If we plug in the values that we are given in the problem, we get:

42 = 7 (a)

To solve for a, we simply divide both sides of the equation by 7.

42/7 = 7a/7

a = 6 m/s^2

Next, we should write out all of the information we have and what we are looking for.

a = 6 m/s^2

v1 = 0 m/s

t = 55 s

v2 = ?

We can use a kinematic equation to solve this problem. We should use:

v2 = v1 + at

If we plug in the values listed above, we should get:

v2 = 0 + (6)(55)

Next, we should solve the problem by performing the multiplication on the right side of the equation.

v2 = 330 m/s

Therefore, the final velocity reached by the object is 330 m/s.

Hope this helps!

7 0

2 years ago

A bird sitting high in a tree is an example of an object with what type of energy?

Wewaii [24]

<span>potential energy because it has the ability to do work

*A*</span>

7 0

2 years ago

Read 2 more answers

A drone is launched with a velocity of 63 m/s, 29 . Three minutes after the drone is launched it suddenly changed its course to

Nadusha1986 [10]

Answer:

v = 127.66 m / s θ’= 337.59

Explanation:

For this exercise we must use the speed composition of the drone.

The first speed is 63 m / s in direction 29, three minutes later the speed reaches 22 m / s and direction 233, finally it returns to the launch point with 98 m / s in direction 321, in the attachment you can see a diagram of these speeds.

To find the resulting average velocity, the easiest thing is to decompose each velocity into the x and y coordinate system, then add each velocity

let's break down the speeds

cos 29 = v₁ₓ / v₁

sin 29 = $v_{1y}$ / v₁

v₁ₓ = v₁ cos 29

v_{1y} = v₁ sin 29

v₁ₓ = 63 cos 29 = 55.10 m / s

v_{1y} = 63 sin 29 = 30.54 m / s

speed 2

cos 22 = v₂ₓ / v₂

sin 22 = v_{2y} / v₂

v₂ₓ = v₂ cos 233

v_{2y} = v₂ sin 233

v₂ₓ = 22 cos 233 = -13.24 m / s

v_{2y} = 22 sin 233 = -17.57 m / s

speed 3

cos 321 = v₃ₓ / v₃

sin 321 = v_{3y} / v₃

v₃ₓ = v₃ cos 321

v_{1y} = vₐ sin 321

v₃ₓ = 98 cos 321 = 76.16 m / s

v_{3y} = 98 sin 321 = -61.67 m / s

We already have all the component of the speeds, the resulting speed is

vₓ = v₁ₓ + v₂ₓ + v₃ₓ

vₓ = 55.10 -13.24 +76.16

vₓ = 118.02 m / s

v_{y} = v_{1y} + v_{2y} + v_{3y}

v_{y} = 30.54 -17.54 - 61.67

v_{y} = -48.67 m / s

there are two ways to give the result

v = (118.02 i -48.67 j) m / s

or in the form of magnitud and angle.

We use the Pythagorean theorem for the module

v = √ (vₓ² + v_{y}²)

v = RA (118.02² + 48.67²)

v = 127.66 m / s

let's use trigonometry for the angle

tan θ = v_{y} / vₓ

θ = tan⁻¹ (v_{1} / vₓ)

θ = tan⁻¹ (-48.67 / 118.02)

θ = -22.41

if we want to measure the angles with respect to the positive side of the x axis

θ’= 360 - 22.41

θ’= 337.59

6 0

2 years ago