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MrRissso [65]
3 years ago
13

There are 22 gloves in a drawer: 5 pairs of red gloves, 4 pairs of yellow, and 2 pairs of green. You select the gloves in the da

rk and can check them only after a selection has been made. What is the smallest number of gloves you need to select to have at least one matching pair in the best case?
Engineering
1 answer:
antiseptic1488 [7]3 years ago
3 0

Answer:

Best case = 2 gloves

Given Information:

Red gloves = 5 pairs

Yellow gloves = 4 pairs

Green gloves = 2 pairs

smallest number of gloves you need to select to have at least one matching pair in the best case = ?

Explanation:

How many gloves do you need to make one matching pair?

2

Yes, you are right. 2 gloves makes a matching pair and it is the smallest number of gloves you need to select to have at least one matching pair.

But what about worst case?

lets say

you tried all 5 red gloves either all of them were left or right

then you tried 4 yellow gloves either all of them were left or right

then you tried 2 green gloves either all of them were left or right

Now all the left or right gloves are tried (5 + 4 + 2 = 11) and the 12th one will definitely be either matching red, yellow or green.

Therefore, in the worst case scenario, the smallest number of gloves you need to select to have at least one matching pair is 12.

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Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters
alex41 [277]

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

 

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

5 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
C#: Arrays - Ask the user how many students names they want to store. You will create two parallel arrays (e.g. 2 arrays with th
zhenek [66]

Answer:

  1. using System;      
  2. public class Program
  3. {
  4. public static void Main()
  5. {
  6.  Console.WriteLine("Enter number of students: ");
  7.  int num = Convert.ToInt32(Console.ReadLine());
  8.  string [] firstName = new string[num];
  9.  string [] lastName = new string[num];
  10.  
  11.  for(int i=0 ; i < num; i++){
  12.   Console.WriteLine("Enter first name: ");
  13.   firstName[i] = Console.ReadLine();
  14.    
  15.   Console.WriteLine("Enter last name: ");
  16.   lastName[i] = Console.ReadLine();
  17.  }
  18.  
  19.  for(int j=0; j < num; j++){
  20.   Console.WriteLine(lastName[j] + "," + firstName[j]);
  21.  }
  22. }
  23. }

Explanation:

Firstly, prompt user to enter number of student to be stored (Line 6- 7). Next, create two array, firstName and lastName with num size (Line 8-9).

Create a for-loop to repeat for num times and prompt user to enter first name and last name and then store them in the firstName and lastName array, respectively (Line 11 - 17).

Create another for loop to traverse through the lastName and firstName array and display the last name and first name by following the format given in the question (Line 19 - 21).

4 0
3 years ago
Guess my birthday it’s in September you should be good with it
Paraphin [41]

Answer:

spetember 7th

Explanation:

5 0
3 years ago
An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 95 kPa and 278C, and 75
Inessa [10]

Answer:

(a). The value of temperature at the end of heat addition process            T_{3} = 2042.56 K

(b). The value of pressure at the end of heat addition process                    P_{3} = 1555.46 k pa

(c). The thermal efficiency of an Otto cycle   E_{otto} = 0.4478

(d). The value of mean effective pressure of the cycle P_{m} = 1506.41 \frac{k pa}{kg}

Explanation:

Compression ratio r_{p} = 8

Initial pressure P_{1} = 95 k pa

Initial temperature T_{1} = 278 °c = 551 K

Final pressure P_{2} = 8 × P_{1} = 8 × 95 = 760 k pa

Final temperature T_{2} = T_{1} × r_{p} ^{\frac{\gamma - 1}{\gamma} }

Final temperature T_{2} = 551 × 8 ^{\frac{1.4 - 1}{1.4} }

Final temperature T_{2} = 998 K

Heat transferred at constant volume Q = 750 \frac{KJ}{kg}

(a). We know that Heat transferred at constant volume Q_{S} = m C_{v} ( T_{3} - T_{2}  )

⇒ 1 × 0.718 × ( T_{3} - 998 ) = 750

⇒ T_{3} = 2042.56 K

This is the value of temperature at the end of heat addition process.

Since heat addition is constant volume process. so for that process pressure is directly proportional to the temperature.

⇒ P ∝ T

⇒ \frac{P_{3} }{P_{2} } = \frac{T_{3} }{T_{2} }

⇒ P_{3} = \frac{2042.56}{998} × 760

⇒ P_{3} = 1555.46 k pa

This is the value of pressure at the end of heat addition process.

(b). Heat rejected from the cycle Q_{R} = m C_{v} ( T_{4} - T_{1}  )

For the compression and expansion process,

⇒ \frac{T_{3} }{T_{2} } = \frac{T_{4} }{T_{1} }

⇒ \frac{2042.56}{998} = \frac{T_{4} }{551}

⇒ T_{4} = 1127.7 K

Heat rejected Q_{R} = 1 × 0.718 × ( 1127.7 - 551)

⇒ Q_{R} = 414.07 \frac{KJ}{kg}

Net heat interaction from the cycle Q_{net} = Q_{S} - Q_{R}

Put the values of Q_{S} & Q_{R}  we get,

⇒ Q_{net} = 750 - 414.07

⇒ Q_{net} = 335.93 \frac{KJ}{kg}

We know that for a cyclic process net heat interaction is equal to net work transfer.

⇒ Q_{net} = W_{net}

⇒ W_{net} = 335.93 \frac{KJ}{kg}

This is the net work output from the cycle.

(c). Thermal efficiency of an Otto cycle is given by

E_{otto} = 1- \frac{T_{1} }{T_{2} }

Put the values of T_{1} & T_{2} in the above formula we get,

E_{otto} = 1- \frac{551 }{998 }

⇒ E_{otto} = 0.4478

This is the thermal efficiency of an Otto cycle.

(d). Mean effective pressure P_{m} :-

We know that mean effective pressure of  the Otto cycle is  given by

P_{m} = \frac{W_{net} }{V_{s} } ---------- (1)

where V_{s} is the swept volume.

V_{s} = V_{1}  - V_{2} ---------- ( 2 )

From ideal gas equation P_{1} V_{1} = m × R × T_{1}

Put all the values in above formula we get,

⇒ 95 × V_{1} = 1 × 0.287 × 551

⇒ V_{1} = 0.6 m^{3}

From the same ideal gas equation

P_{2} V_{2} = m × R × T_{2}

⇒ 760 × V_{2} = 1 × 0.287 × 998

⇒ V_{2} = 0.377 m^{3}

Thus swept volume V_{s} = 0.6 - 0.377

⇒ V_{s} = 0.223 m^{3}

Thus from equation 1 the mean effective pressure

⇒ P_{m} = \frac{335.93}{0.223}

⇒ P_{m} = 1506.41 \frac{k pa}{kg}

This is the value of mean effective pressure of the cycle.

4 0
3 years ago
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