The CountLetters application is limited to counting letters in a single word. Modify the CountLetters application to count the letters in an entire phrase, which contains spaces. Care must be taken to ignore the spaces and any other non alphabetic character found in the phrase. Be sure to change comments and variable names appropriately so that the reader of the application code understands that the letters in a phrase are counted.


import java.util.Scanner;

public class CountLetters {

public static void main(String[] args) {
final int LOW = 'A'; //smallest possible value
final int HIGH = 'Z'; //highest possible value
int[] letterCounts = new int[HIGH - LOW + 1];
Scanner input = new Scanner(System.in);
String phrase;
char[] wordLetters;
int offset; //array index

/* prompt user for a phrase */
System.out.print("Enter a phrase: ");
phrase = input.nextLine();
input.close();

/* convert word to char array and count letter occurrences */
phrase = phrase.toUpperCase();
wordLetters = phrase.toCharArray();
for (int letter = 0; letter < wordLetters.length; letter++) {
offset = wordLetters[letter] - LOW;
letterCounts[offset] += 1;
}

/* show letter occurrences */
for (int i = LOW; i <= HIGH; i++) {
System.out.println((char)i + ": " + letterCounts[i - LOW]);
}
}

}

Answer:

I dont really know, I am sorry, but I am going to ask my teacher

Answer: green

Explanation:

Airbags are supplemental protection and are not designed to deploy in all crashes. Therefore, it's false.

It should be noted that airbags are out in vehicles in order to reduce the impact on an individual during accidents.

Therefore, airbags are supplemental protection and are not designed to deploy in all crashes. They're designed to work with seatbelts.

Learn more about airbag on:

brainly.com/question/2607849

#SPJ9

Answer:

C47 may seem like a fancy word, but in the film world, it is the name of one of the simplest, most useful and versatile tools, which is commonly knows as a clothespin.

Following are major uses of C47 on a film set.

1) Lightning Purposes

C47 are used for attaching materials like gel and diffusion to the adjustable flaps, that control the direction of light

2) Adjusting Wardrobes

C47 are used to shorten, tighten or redesign a material of clothing for a specific purpose. It avoids permanent change and the hassle of sewing.

3) Modifications

These can be used for temporary repairs or to modify the set and shooting locations, e.g if a curtain is not hanging properly, C47 can be used

Answer:

The correct option is;

$10,000 of Personal Injury Protection (PIP)

Explanation:

Personal injury protection, PIP, otherwise known as No-Fault Law coverage is a part of auto insurance that takes care of medical cost of treatment of an accident victim regardless of who is at fault

In Florida, the No-Fault Law requires motor vehicle owners to have and retain PIP providing $10,000 in medical care, resulting disability and expenses of a funeral resulting from an accident in which the motorist is involved

Therefore, the correct option is $10,000 of Personal Injury Protection.

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, [tex]T_C[/tex]  = -15°C = 258.15 K

Temperature of the cooling water, [tex]T_H[/tex] = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

[tex]\gamma_{Max} = \dfrac{T_C}{T_H - T_C} = \dfrac{258.15}{303.15 - 258.15} = 5.74[/tex]

(2) For ammonia refrigerant, we have;

[tex]h_2 = h_g = 1466.3 \ kJ/kg[/tex]

[tex]h_3 = h_f = 322.42 \ kJ/kg[/tex]

[tex]h_4 = h_3 = h_f = 322.42 \ kJ/kg[/tex]

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

[tex]h_1 = h_{f1} + x_1 \times h_{gf}[/tex]

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

[tex]\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}[/tex]

[tex]\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09[/tex]

(3) The circulation rate is given by the mass flow rate, [tex]\dot m[/tex] as follows

[tex]\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}[/tex]

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

[tex]\dot m = \dfrac{105}{956.08} = 0.1098 \ kg/h[/tex]

[tex]\dot m[/tex] = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×[tex]\dot m[/tex]

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

Answer: you are reducing bullet damage or the damage of a person getting shattered glass all over them

Explanation: i dont

have one

Answer:

0.635 m

Explanation:

When calculating the resistance R of a wire, we need  its length(l), its cross-sectional area (A) and the resistivity of the material(ρ). The resistance of a wire  is given by the equation:

Resistance (R) = Resistivity(ρ) × length (l) / cross-sectional area (A)

For a 12 guage wire,

Resistivity (ρ) = 1.724 × 10 ⁻⁸ ohm m, length (l) = 400 ft = 121.92 m,

Diameter (d) = 0.00205232, cross-sectional area (A) = πd²/4 = π(0.00205232)²/4 = 3.31 × 10 ⁻⁶ m²

[tex]R=\frac{\rho l}{A}=\frac{1.724*10^{-8}*121.92}{3.31*10^{-6}} =0.635m[/tex]