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 option that is not associated with the given term called urban sprawl is; Option A: Blocking high views
What is Urban Sprawl?
Urban sprawl is defined as the rapid expansion of the geographic boundaries of towns and cities which is often accompanied by low-density residential housing and increased reliance on the private automobilefor movement.
Looking at the given options, "blocking high views" is the option that is not typically a problem associated with urban sprawl because urbanization usually takes place on relatively flat levels.
The missing options are;
a. blocking high views
b. destroying animal habitats
c. overrunning farmland
d. reducing green space
Read more about urban sprawl at; brainly.com/question/504389
Answer:
class TriangleNumbers
{
public static void main (String[] args)
{
for (int number = 1; number <= 10; ++number) {
int sum = 1;
System.out.print("1");
for (int summed = 2; summed <= number; ++summed) {
sum += summed;
System.out.print(" + " + Integer.toString(summed));
}
System.out.print(" = " + Integer.toString(sum) + '\n');
}
}
}
Explanation:
We need to run the code for each of the 10 lines. Each time we sum numbers from 1 to n. We start with 1, then add numbers from 2 to n (and print the operation). At the end, we always print the equals sign, the sum and a newline character.
Answer:
can you please ask in English I can't understand this language
Ik i am thank you tho xoxo