Answer:
The maximum theoretical height that the pump can be placed above liquid level is 
Explanation:
To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:
(
stands here for density, 
for height)
Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:
This means that pressure drop is proportional to the suction lift's height.
We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.
That means:
We insert that into our last equation and get:
And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.
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;
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A formal indication from a state, on letterhead, or an official state form, which shows that an applicant has valid driving privileges and is clear to apply for a Colorado driver's license is called a clearance.
<h3>What is a learner's license?</h3>
A learner's license is also referred to as learner's permit and it can be defined as a category of driver's license that is issued to an individual who is learning how to drive an automobile vehicle (car).
<h3>What is a license?</h3>
A license is also referred to as a certificate and it can be defined as an authorization that is typically issued by state governments to a driver, so as to avail him or her the legal ability to physically driver in all the states across a country.
In this scenario, we can infer and logically deduce that a clearance is a formal indication from a state which shows and affirms that an applicant has valid driving privileges, and is permitted to apply for a Colorado driver's license.
Learn more about learner's license here: brainly.com/question/26289148
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Answer:
the maximum bending stress in the strap is 3.02 ksi
Explanation:
Given the data in the question;
steel strap thickness = 0.125 in
width = 2 in
circular arc radius = 600 in
we know that, standard value of modulus of elasticity of L2 steel is; E = 29 × 10³ ksi;
Now, using simple theory of bending
1/p = M/EI
solve for M
Mp = EI
M = EI / p ----- let this be equation 1
The maximum bending stress in the strap is;
σ = Mc / I -------let this be equation 2
substitute equation 1 into 2
σ = ( EI / p)c / I
σ = ( c/p )E
so we substitute in our values
σ = ( (0.125/2) / 600 )29 × 10³
σ = 0.00010416666 × 29 × 10³
σ = 3.02 ksi
Therefore, the maximum bending stress in the strap is 3.02 ksi
