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Tpy6a [65]
2 years ago
11

The zero tolerance law applies to drivers _________.

Physics
1 answer:
s2008m [1.1K]2 years ago
4 0

The zero tolerance law applies to drivers <u>under the age of 21.</u>

<h3>Zero Tolerance Law:</h3>

Drivers under the age of 21 who operate a motor vehicle with a BAC of between 0.02% and 0.07% are subject to what is known as the "Zero Tolerance Law." The Zero Tolerance Law aims to prevent underage drivers from driving after consuming alcohol.

The zero-tolerance rule states that it is against the law for anyone under the age of 21 to operate a motor vehicle while having any detectable level of alcohol in their system. Zero Tolerance Law states that a minor has committed the criminal offense of DUI if there is ANY detectable level of alcohol in his or her system while operating a vehicle in public. The minor's driver's license is immediately suspended, and the officer has the authority to confiscate the license on the spot.

Learn more about zero tolerance here:

brainly.com/question/17257160

#SPJ4

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Which characeristic makes omasis different from diffusion.
Ksju [112]
 <span>haha I used to think biology was so hard, i find it quite easy now. 
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OSMOSIS, the important thing to remember is that water ALWAYS flow towards the region with the higher concentration of the solute (ex: Salt is solute, water is solvent) solute is the thing that is being dissolved. Solvent is the one doing the dissolving. Hope this helped!</span>
5 0
3 years ago
What experimental evidence led to the development of this atomic model from the one before it?
Marina86 [1]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back

Explanation:

7 0
3 years ago
Read 2 more answers
Suppose Earth's gravitational force were decreased by half. How would this change affect a game of basketball? Write a paragraph
timama [110]
If the gravitational force were<span> decreased by half, there would be lack of gravity on earth. Hence, it would basically affect the velocity, speed, and the distance travelled in any direction by basketball players and the ball. The basketball would bounce higher and come down in a slower speed. Whereas for the players, they would be able to leap higher from the floor.</span><span> </span>
7 0
3 years ago
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0
3 years ago
Use the equation PiVi PfVf. Assume that Pi 101 kPa and Vi 10.0 L. If Pf 43.0 kPa, what is Vf
iren2701 [21]
Solving for vf gives you PiVi/Pf. Now plug in 101kPa*10L/43kPa = 23.48L. Using significant figures i would round to 23.5L
8 0
3 years ago
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