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n200080 [17]
3 years ago
12

A pump transfers water from a lake to a reservoir, which is located 29.2 m above the lake, at a rate of 11.5 L/s. Determine the

amount of power (in kW) required to pump water from the lake to the reservoir.
Engineering
1 answer:
Greeley [361]3 years ago
8 0

Answer:

P=3290.84KW

Explanation:

From the question we are told that:

Heighth=29.2m

Rate Q=11.5L/s=>11.5*10^{-3}

Generally the equation for Power is mathematically given by

P=\rho*g*Q*h

P=1000*9.81*11.5*10^{-3}*29.2

P=3290.84KW

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Write two scnr.nextInt statements to get input values into birthMonth and birthYear. Then write a statement to output the month,
aalyn [17]

Answer:

import java.util.Scanner;

public class InputExample {

   public static void main(String[] args) {

       Scanner scnr = new Scanner(System.in);

       int birthMonth;

       int birthYear;

       birthMonth = scnr.nextInt();

       birthYear = scnr.nextInt();

       System.out.println(birthMonth+"/"+birthYear);

   }

}

3 0
3 years ago
The mechanical energy of an object is a combination of its potential energy and its
saveliy_v [14]

The mechanical energy of an object is a combination of its potential energy and its <em><u>kinetic</u></em><em><u> </u></em><em><u>energy</u></em><em><u>.</u></em>

6 0
2 years ago
A 3 m aluminum pole is kept at a residential site for construction
Aliun [14]

Answer:

I don't know sorry

Explanation:

5 0
3 years ago
2. The moist weight of 0.1 ft3 of soil is 12.2 lb. If the moisture content is 12% and the specific gravity of soil solids is 2.7
adell [148]

The answers to dry unit weight, void ratio, porosity, degree of saturation, volume occupied by water are respectively;

γ_d = 108.93 lb/ft³; e = 0.56; n = 0.36; S = 0.58; V_w = 0.021 ft³

<h3>Calculation of Volume and Weight of soil</h3>

We are given;

Moist weight; W = 12.2 lb

Volume of moist soil; V = 0.1 ft³

moisture content; w = 12% = 0.12

Specific gravity of soil solids; G_s = 2.72

A) Formula for dry unit weight is;

γ_d = γ/(1 + w)

where γ_w is moist unit weight as;

γ_w = W/V

γ_w = 122/0.1 = 122 lb/ft³

Thus;

γ_d = 122/(1 + 0.12)

γ_d = 108.93 lb/ft³

B) Formula for void ratio is;

e = [(G_s * γ_w)/γ_d] - 1

e = [(2.72 * 122)/108.93] - 1

e = 0.56

C) Formula for porosity is;

n = e/(1 + e)

n = 0.56/(1 + 0.56)

n = 0.36

D) Formula for degree of saturation is;

S = (w * G_s)/e

S = (0.12 * 2.72)/0.56

S = 0.58

E) Volume occupied by water is gotten from;

V_w = S*V_v

where;

V_v is volume of voids = nV

V_v = 0.36*0.1

V_v = 0.036 ft³

Thus;

V_w = 0.58 * 0.036

V_w = 0.021 ft³

Read more about Specific Gravity of Soil at; brainly.com/question/14932758

4 0
3 years ago
Suppose you have two boxes in front of you. One box contains a Thevenin Equivalent (voltage source in series with a resistor) an
fomenos

Answer:

1. Measure the temperature of the boxes and leave them unconnected.

2. Norton reduces his circuit down to a single resistance in parallel with a constant current source. A real-life Norton equivalent circuit would be continuously wasting power (as heat) as the current source dumps energy into the resistor, even when externally unconnected, while a Thevenin equivalent circuit would sit there doing nothing.

3. The Norton equivalent box would get warm and eventually run out of power. The Thevenin equivalent box would stay at ambient temperature.

8 0
3 years ago
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