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navik [9.2K]
4 years ago
14

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

Physics
2 answers:
lesya [120]4 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]4 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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If a dog is mass iS 14.3 kg what is it’s weight on earth
BigorU [14]

Answer:

Weight of the dog on surface of earth is 140.14 Newton.

Given:

mass of the dog = 14.3 kg

To find:

Weight of the dog = ?

Formula used:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog

g = acceleration due to gravity

Solution:

Weight of the dog is given by,

W = mg

Where, W = weight of the dog

m = mass of the dog = 14.3 kg

g = acceleration due to gravity

W = 14.3 × 9.8

W = 140.14 Newton

Weight of the dog on surface of earth is 140.14 Newton.


8 0
3 years ago
In midair an M = 145 kg bomb explodes into two pieces of m1 = 115 kg and another, respectively. Before the explosion, the bomb w
Daniel [21]

Answer:

v_2=-133.17m/s, the minus meaning west.

Explanation:

We know that linear momentum must be conserved, so it will be the same before (p_i) and after (p_f) the explosion. We will take the east direction as positive.

Before the explosion we have p_i=m_iv_i=Mv_i.

After the explosion we have pieces 1 and 2, so p_f=m_1v_1+m_2v_2.

These equations must be vectorial but since we look at the instants before and after the explosions and the bomb fragments in only 2 pieces the problem can be simplified in one dimension with direction east-west.

Since we know momentum must be conserved we have:

Mv_i=m_1v_1+m_2v_2

Which means (since we want v_2 and M=m_1+m_2):

v_2=\frac{Mv_i-m_1v_1}{m_2}=\frac{Mv_i-m_1v_1}{M-m_1}

So for our values we have:

v_2=\frac{(145kg)(24m/s)-(115kg)(65m/s)}{(145kg-115kg)}=-133.17m/s

5 0
3 years ago
I WILL MARK YOU THE BRAINLIEST NO LINKS
nasty-shy [4]

yeah it definitely 2 .:) :)

3 0
3 years ago
john doe operates a crane that can pick up 3.0 tons of excavated earth in an hour. John's wages are $35 per hour. What, then, is
levacccp [35]

Answer:

1

Explanation:

8 0
3 years ago
The electron affinity of thulium has been measured by a technique known as laser photodetachment electron spectroscopy. In this
antoniya [11.8K]

Answer:

ΔE = 1.031 eV

Explanation:

For this exercise let's calculate the energy of the photons using Planck's equation

          E = h f

wavelength and frequency are related

         c = λ f

         f = c /λ

let's substitute

         E = h c /λ

let's calculate

         E = 6.63 10⁻³⁴ 3 10⁸/1064 10⁻⁹

         E = 1.869 10⁻¹⁹ J

let's reduce to eV

         E = 1.869 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

        E = 1.168 eV

therefore the electron affinity is

         ΔE = E - 0.137

         ΔE = 1.168 - 0.137

         ΔE = 1.031 eV

3 0
3 years ago
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