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Firlakuza [10]
2 years ago
8

____________ is the range of all the colors created by different amounts of light.

Engineering
2 answers:
astra-53 [7]2 years ago
7 0
The answer to your question is value.
morpeh [17]2 years ago
6 0
I think the answer is constrast
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A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr
Hoochie [10]

Answer:

\Delta m = 102.25\,kg

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

m_{o} = \rho_{air}\cdot V_{tank}

m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{o} = 31.25\,kg

The final mass inside the rigid tank is:

m_{f} = \rho \cdot V_{tank}

m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{f}= 133.5\,kg

The supplied air mass is:

\Delta m = m_{f}-m_{o}

\Delta m = 133.5\,kg-31.25\,kg

\Delta m = 102.25\,kg

4 0
3 years ago
Write a program that removes all spaces from the given input. You may assume that the input string will not exceed 50 characters
GrogVix [38]

Answer:

Program that removes all spaces from the given input

Explanation:

// An efficient Java program to remove all spaces  

// from a string  

class GFG  

{  

 

// Function to remove all spaces  

// from a given string  

static int removeSpaces(char []str)  

{  

   // To keep track of non-space character count  

   int count = 0;  

 

   // Traverse the given string.  

   // If current character  

   // is not space, then place  

   // it at index 'count++'  

   for (int i = 0; i<str.length; i++)  

       if (str[i] != ' ')  

           str[count++] = str[i]; // here count is  

                                   // incremented  

         

   return count;  

}  

 

// Driver code  

public static void main(String[] args)  

{  

   char str[] = "g eeks for ge eeks ".toCharArray();  

   int i = removeSpaces(str);  

   System.out.println(String.valueOf(str).subSequence(0, i));  

}  

}  

5 0
3 years ago
Read 2 more answers
A cylinder with a frictionless piston contains 0.05 m3 of air at 60kPa. The linear spring holding the piston is in tension. The
AleksAgata [21]

Answer:

18 kJ

Explanation:

Given:

Initial volume of air = 0.05 m³

Initial pressure = 60 kPa

Final volume = 0.2 m³

Final pressure = 180 kPa

Now,

the Work done by air will be calculated as:

Work Done = Average pressure × Change in volume

thus,

Average pressure = \frac{60+180}{2}  = 120 kPa

and,

Change in volume = Final volume - Initial Volume = 0.2 - 0.05 = 0.15 m³

Therefore,

the work done = 120 × 0.15 = 18 kJ

4 0
3 years ago
A building wall has dimensions of 3 m tall and 10 m wide. It is constructed of 2 cm. wallboard (k = 0.5 W/m-C) on the inside, 3
Art [367]

Answer: heat loss through wall is 16.58034kW

Temperature of inside wall surface is 47°c

Temperature of outside wall surface is -2.7°c

Explanation:detailed calculation and explanation is shown in the image below.

4 0
3 years ago
When a retaining structure moves towards the soil backfill, the stress condition is called:__________.
Alecsey [184]

Answer:

(C) passive state.

Explanation:

The earth pressure is the pressure exerted by the soil on the shoring system. They are three types of earth pressure which are:

a) Rest state: In this state, the retaining wall is stationary, this makes the lateral stress to be zero.

b) Active state: In this state, the wall moves away from the back fill, this leads to an internal resistance. Hence the active earth pressure is less than earth pressure at rest

c) Passive state: In this state the wall is pushed towards the back fill, this leads to shearing resistance. Hence, the passive earth pressure is greater than earth pressure at rest

6 0
2 years ago
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