I NEED HELP ASAP PLEASE!!!!!

The zero tolerance law applies to drivers _________.

s2008m [1.1K]

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:

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4 0

2 years ago

A 1.2kg ball rolls forward with an acceleration of 1.11 m/s. What is the net force on the ball

grandymaker [24]

Answer:

1.332 N

Explanation:

Net Force = Mass x Acceleration
1.2 x 1.11 = 1.332 N

I'm so sorry if I'm wrong.

8 0

1 year ago

A 20 kg mass is dropped from a tall rooftop and accelerates at 9.8 m/s2. What is the weight of the dropped object?

patriot [66]

Objects in free fall are weightless.

6 0

3 years ago

Read 2 more answers

A physics teacher performing an outdoor demonstration suddenly falls from rest off a high cliff and simultaneously shouts "Help"

Digiron [165]

Answer: a) The cliff is 532.05m high

b) Her speed just before hitting the ground is 102.12 m/s

Explanation: To solve This, I'll use a sketch diagram, attached to this solution,

In 3seconds, the teacher heard the echo of her initial scream back. We can obtain the distance the teacher had fallen at the end of 3 seconds using the equations of motion,

Y1 = ut + 0.5g(t^2)

Since she's falling under the influence of gravity, her initial velocity, u = 0m/s, g = 9.8m/s2, t = 3s

Y1, distance she fell through in 3 seconds = 0.5×9.8(3^2) = 44.1m

Let the total height of the cliff be (44.1 + x); where is the remaining height of cliff that the teacher will fall through.

Using the equations of motion again, we can obtain distance travelled by the sound waves in 3s. sound waves travel with a constant speed of 340m/s, no acceleration,

Y2 = ut + 0.5g(t^2) where g = 0, u = 340m/s, t = 3seconds

Y2 = 340 × 3 = 1020m

But in 3 secs, the sound waves would have travelled through the total height of the cliff (44.1 + x) and back to the teacher's current height, x. That is, 1020 = 44.1 + x + x

x = 487.95m

So, total height of cliff = 44.1 + 487.95 = 532.05m

b) the speed of the teacher just before she hits the ground.

Using the equations of motion again,

(V^2) = (U^2) + 2gs

Where v is the final velocity to be calculated

U is the initial velocity = 0m/s

g is acceleration due to gravity = 9.8m/s2

S is the total height she fell through, that is, the height of the cliff = 532.05m

(V^2) = 0 + 2×9.8×532.05 = 10428.18

V = √(10428.18) = 102.12m/s

QED!

4 0

3 years ago

A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin

emmainna [20.7K]

(1)

Cheetah speed: $v_c = 97.8 km/h=27.2 m/s$

Its position at time t is given by

$S_c (t)= v_c t$ (1)

Gazelle speed: $v_g = 78.2 km/h=21.7 m/s$

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

$S_g(t)=S_0 +v_g t$ (2)

The cheetah reaches the gazelle when $S_c=S_g$ . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle: