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gulaghasi [49]
2 years ago
14

The Hoover Dam (see the Video) backs up the Colorado River and creates Lake Meade, which is approximately 115 miles long and has

a surface area of approximately 225 square miles. If during flood conditions the Colorado River flows into the lake at a rate of 43800 cfs and the outflow from the dam is 6200 cfs, how many feet per 24-hr day will the lake level rise?
Engineering
1 answer:
Kamila [148]2 years ago
7 0

Answer: 0.518

Explanation: Total Area of lake = 225 sq miles = 6.273e+9 sq feet

Change in volume per second = 43800-6200 = 37600 cfs

Change in volume per day = 37600 * 3600 * 24 = 3.249e+9 cfd

Total rise of water level per day  = Change in volume per day/Total lake area

                                                       = 3.249e+9/6.273E+9 = 0.518 feet/day

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A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting
sergij07 [2.7K]

Answer:

v₀ = 2,562 m / s  = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

              F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

              F dx = m v dv

We replace and integrate

            -β ∫ x³ dx = m ∫ v dv

            β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

        -β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

         x_max⁴ = 2 m /β   v₀²

         

Let's look for the speed that the train can have for maximum compression

         x_max = 20 cm = 0.20 m

         

         v₀ =√(β/2m)   x_max²

Let's calculate

          v₀ = √(640 106/2 7.8 104)    0.20²

          v₀ = 64.05  0.04

          v₀ = 2,562 m / s

          v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

          v₀ = 9.2 km / h

5 0
2 years ago
A distância entre duas retas reversas é a medida de um segmento orientado que *?
Agata [3.3K]

Answer:

Fatec – SP) Seja A um ponto pertencente à reta r, contida no plano α. É verdade que:

a) existe uma única reta que é perpendicular à reta r no ponto A.

b) existe uma única reta, não contida no plano α, que é paralela à reta r.

c) existem infinitos planos distintos entre si, paralelos ao plano α, que contém a reta r.

d) existem infinitos planos distintos entre si, perpendiculares ao plano α e que contêm a reta r.

e) existem infinitas retas distintas entre si, contidas no plano α e que são paralelas à reta r.

2. (UF – AL) Classifique como verdadeira ou falsa cada uma das afirmativas abaixo.

1) Duas retas que não têm pontos com

8 0
2 years ago
Fix the code so the program will run correctly for MAXCHEESE values of 0 to 20 (inclusive). Note that the value of MAXCHEESE is
GarryVolchara [31]

Answer:

Code fixed below using Java

Explanation:

<u>Error.java </u>

import java.util.Random;

public class Error {

   public static void main(String[] args) {

       final int MAXCHEESE = 10;

       String[] names = new String[MAXCHEESE];

       double[] prices = new double[MAXCHEESE];

       double[] amounts = new double[MAXCHEESE];

       // Three Special Cheeses

       names[0] = "Humboldt Fog";

       prices[0] = 25.00;

       names[1] = "Red Hawk";

       prices[1] = 40.50;

       names[2] = "Teleme";

       prices[2] = 17.25;

       System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");

       System.out.println(names[0] + ": $" + prices[0] + " per pound");

       System.out.println(names[1] + ": $" + prices[1] + " per pound");

       System.out.println(names[2] + ": $" + prices[2] + " per pound");

       Random ranGen = new Random(100);

       // error at initialising i

       // i should be from 0 to MAXCHEESE value

       for (int i = 0; i < MAXCHEESE; i++) {

           names[i] = "Cheese Type " + (char) ('A' + i);

           prices[i] = ranGen.nextInt(1000) / 100.0;

           amounts[i] = 0;

           System.out.println(names[i] + ": $" + prices[i] + " per pound");

       }        

   }

}

7 0
2 years ago
. A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa a
Delvig [45]

Answer:

Qin = 1857 kJ

Explanation:

Given

m = 0.5 Kg

T₁ = 25°C = (25 + 273) K = 298 K

P₁ = 100 kPa

P₂ = 500 kPa

First, the temperature when the piston starts rising is determined from the ideal gas equations at the initial state and at that state:

T₂ = T₁*P₂/P₁

⇒  T₂ = 298 K*(500 kPa/100 kPa) = 1490 K

Until the piston starts rising no work is done so the heat transfer is the change in internal energy

Qin = ΔU = m*cv*(T₂-T₁)

⇒   Qin = 0.5*3.1156*(1490 - 298) kJ = 1857 kJ

8 0
2 years ago
A cylindrical rod of copper (E = 110 GPa, 16 × 106 psi) having a yield strength of 240 MPa (35,000 psi) is to be subjected to a
77julia77 [94]

Answer:

7.65 mm

Explanation:

Stress, \sigma=\frac {F}{A} where F is the force and A is the area

Also, \sigma=E\times \frac {\triangle L}{L}

Where E is Young’s modulus, L is the length and \triangle L is the elongation

Therefore,

\frac {F}{A}= E\times \frac {\triangle L}{L}

Making A the subject of the formula then

A=\frac {FL}{E\triangle L}=\frac {6660\times 380}{110\times 10^{9}\times 0.5}=4.60145\times 10^{-5} m^{2}

Since A=\frac {\pi d^{2}}{4} then  

d=\sqrt {\frac {4A}{\pi}}=\sqrt {\frac {4\times 4.60145\times 10^{-5}}{\pi}}= 0.00765425m= 7.654249728 mm\approx 7.65 mm

4 0
2 years ago
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