Answer:
V = 0.5 m/s
Explanation:
given data:
width of channel = 4 m
depth of channel = 2 m
mass flow rate = 4000 kg/s = 4 m3/s
we know that mass flow rate is given as
Putting all the value to get the velocity of the flow
V = 0.5 m/s
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
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Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
Technician B
Explanation:
Technician B is correct in his argument. This is because according to what he said, as the computer pulses stimuli the coil will turn on and off, promoting an increase in the voltage that will cause the fluctuation. Technician A is incorrect because the procedure he indicated imposes that the voltage is checked at the negative terminal and not at the positive.
Answer:
It will be B
Explanation:
Since resisitors in series are added together, 1 + 1 + 1 would = 3kilo ohms. But with resistors in parallel would be (1/1+1/1)^-1. That would equal 0.5 Now you have two resistors in series for B, and because now that they are in series you add them together, so 0.5 + 1 = 1.5 kilo ohms which is what is needed.
