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mel-nik [20]
2 years ago
8

Drivers education - Unit 3

Engineering
1 answer:
melamori03 [73]2 years ago
4 0

The following scenarios are pertinent to driving conditions that one may encounter. See the following rules of driving.

<h3>What do you do when the car is forced into the guardrail?</h3>

Best response:

  • I'll keep my hands on the wheel and slow down gradually.
  • The reason I keep my hands on the steering wheel is to avoid losing control.
  • This will allow me to slowly back away from the guard rail.
  • The next phase is to gradually return to the fast lane.
  • Slamming on the brakes at this moment would result in a collision with the car behind.

Scenario 2: When driving on a wet road and the car begins to slide

Best response:

  • It is not advised to accelerate.
  • Pumping the brakes is not recommended.
  • Even lightly depressing and holding down the brake pedal is not recommended.
  • The best thing to do is take one foot off the gas pedal.
  • There should be no severe twists at this time.

Scenario 3: When you are in slow traffic and you hear the siren of an ambulance behind

Best response:

  • The best thing to do at this moment is to go to the right side of the lane and come to a complete stop.
  • This helps to keep the patient in the ambulance alive.
  • It also provide a clear path for the ambulance.
  • Moving to the left is NOT recommended.
  • This will exacerbate the situation. If there is no place to park on the right shoulder of the road, it is preferable to stay in the lane.

Learn more about rules of driving. at;

brainly.com/question/8384066

#SPJ1

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Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.
marin [14]

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30  + ( bicarbonate ion ) * 50/61

                                   = 15 * (50/30) + 120*( 50/61 )  = 123.361 mg/l  as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57  = 2.692*10^-5  moles/l

[ OH^- ]   = 2.692*10^-5  * 179/mole * 10^3 mg/g  = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3  = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

8 0
3 years ago
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