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Vesnalui [34]
3 years ago
8

Two AAA-size lithium batteries are connected in series in a flashlight. Each battery has 3.5 volt and 4- Amp-hour capacity. If t

he flashlight is connected to a light bulb with 10 Ohms resistance. How long does it take to drain the batteries?
Engineering
1 answer:
mafiozo [28]3 years ago
7 0

Answer:

t= 9.79 hr

Explanation:

Given that

V= 3.5 V

Capacity= 4 Amp-hour

We know that

V= IR

V= Voltage

I =Current

R=Resistance

V = I R

The total voltage on the  batteries will be  2 V

2 x 3.5 = I x 10

I= 0.7 A

We know that Power P

P = V I

P = 0.7 x 7

P =4.9 W

4 A.h and 12 volt power supply = 4 x 12 = 48 W.hr

So time of drain t

4.9 t = 48

t= 9.79 hr

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4.68 Steam enters a turbine in a vapor power plant operating at steady state at 560°C, 80 bar, and exits as a saturated vapor at
garik1379 [7]

Answer:

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Explanation:

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3 0
2 years ago
Which traits are common in all four career pathways of the Information Technology field? Check all that apply.
Komok [63]

Answer:

Accuracy and attention to detail, Problem solving and critical thinking skills, Knowledge of programming language .

Explanation:

It is a technological area in which a person learns how to develop computer hardware, including PCs, laptops, tablets, processing, networking, and other hardware parts. Another field of study in IT is Management Information Systems (MIS).

The IT industry's career paths can be categorized equally in the two primary field’s hardware and software areas  

In hardware, there is Production, maintenance, research and development, and strategic planning.

In software, there is manufacturing, development, programming, software testing, and maintenance and support under software.  

Computer operations, administration of databases, sales / marketing and data centre management are connected areas.

7 0
2 years ago
2. Write a Java program that generates a new string by concatenating the reversed substrings of even indexes and odd indexes sep
Nana76 [90]

Answer:

  1. public class Main {
  2.    public static void main(String[] args) {
  3.        String testString = "abscacd";
  4.        String evenStr = "";
  5.        String oddStr = "";
  6.        for(int i=testString.length() - 1; i >= 0; i--){
  7.            if(i % 2 == 0){
  8.                evenStr += testString.charAt(i);
  9.            }
  10.            else{
  11.                oddStr += testString.charAt(i);
  12.            }
  13.        }
  14.        System.out.println(evenStr + oddStr);
  15.    }
  16. }

Explanation:

Firstly, let declare a variable testString to hold an input string "abscacd" (Line 1).

Next create another two String variable, evenStr and oddStr and initialize them with empty string (Line 5-6). These two variables will be used to hold the string at even index and odd index, respectively.

Next, we create a for loop that traverse the characters of the input string from the back by setting initial position index i to  testString.length() - 1  (Line 8). Within the for-loop, create if and else block to check if the current index, i is divisible by 2, (i % 2 == 0), use the current i to get the character of the testString and join it with evenStr. Otherwise, join it with oddStr (Line 10 -14).

At last, we print the concatenated evenStr and oddStr (Line 18).  

4 0
3 years ago
A body weighs 50 N and hangs from a spring with spring constant of 50 N/m. A dashpot is attached to the body. If the body is rai
lbvjy [14]

Answer:

a) 3.607 m

b) 1.5963 m

Explanation:

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3 0
3 years ago
Here, we want to become proficient at changing units so that we can perform calculations as needed. The basic heat transfer equa
netineya [11]

Answer:

9500 kJ; 9000 Btu

Explanation:

Data:

m = 100 lb

T₁ = 25 °C

T₂ = 75 °C

Calculations:

1. Energy in kilojoules

ΔT = 75 °C - 25 °C = 50 °C  = 50 K

m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}

2. Energy in British thermal units

\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}

7 0
3 years ago
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