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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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What causes long sightedness?how is it corrected.

NNADVOKAT [17]

Answer:

Long sight occurs when the eyeball is too short or the lens is too thin, or both. As a result, light rays from near objects are focused behind the retina because the light rays are not converged enough. The image formed on the retina is therefore out of focus.

To correct this problem, people can wear glasses with convex lenses. Light rays from near objects are converged by the convex lenses before entering the eyes, so that light can be focused on the retina to form a sharp image. Additionally, long sight can also be corrected by surgical methods such as LASIK.

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3 years ago

Unpolarized light is incident upon two ideal polarizing filters that do not have their transmission axes aligned. If 19% of the

Zepler [3.9K]

Answer:

51.94°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.19I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.19I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.19I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.19I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.19\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.38}\\\Rightarrow \theta=51.94^{\circ}$

The angle between the two filters is 51.94°

5 0

2 years ago

Is 5m.what<br>In si system, the length of a body is<br>does it mean?

lozanna [386]

Answer: Some conversions from one system of units to another need to be exact, without increasing or decreasing the precision of the first measurement. This is sometimes called soft conversion. It does not involve changing the physical configuration of the item being measured.

Explanation:

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3 years ago

Help needed

MakcuM [25]

Answer: Atmosphere and geosphere.

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6 0

3 years ago

A spring of constant 20 N/m has compressed a distance 8 m by a(n) 0.3 kg mass, then released. It skids over a frictional surface

Fiesta28 [93]

Answer:

$X_2=25.27m$

Explanation:

Here we will call:

1. $E_1$ : The energy when the first spring is compress

2. $E_2$ : The energy after the mass is liberated by the spring

3. $E_3$ : The energy before the second string catch the mass

4. $E_4$ : The energy when the second sping compressed

so, the law of the conservations of energy says that:

1. $E_1 = E_2$

2. $E_2 -E_3= W_f$

3. $E_3 = E_4$

where $W_f$ is the work of the friction.

1. equation 1 is equal to:

$\frac{1}{2}Kx^2 = \frac{1}{2}MV_2^2$

where K is the constant of the spring, x is the distance compressed, M is the mass and $V_2$ the velocity, so:

$\frac{1}{2}(20)(8)^2 = \frac{1}{2}(0.3)V_2^2$

Solving for velocity, we get:

$V_2$ = 65.319 m/s

2. Now, equation 2 is equal to:

$\frac{1}{2}MV_2^2-\frac{1}{2}MV_3^2 = U_kNd$

where M is the mass, $V_2$ the velocity in the situation 2, $V_3$ is the velocity in the situation 3, $U_k$ is the coefficient of the friction, N the normal force and d the distance, so: