Answer:
Explanation:
It is the voltage a voltmeter would read when connected across something that has resistance.
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The diagram above is supposed to represent 3 lightbulbs connected in series. The vertical lines in the middle are supposed to be a battery which powers the three light bulbs. If you put a voltmeter across one of the lightbulbs, it will read a voltage that is 1/3 of the voltage of the battery.
Answer
That reading you get across the one light bulb is The Voltage Drop.
Answer:that is associated with diarrhea in humans. ... Examples of giardia in a Sentence ... These example sentences are selected automatically from various online news sources to reflect current usage of the word 'giardia.
Explanation:
Answer:
Read the passage. Then, answer the questions about the metaphor in boldface in the text.
Lately, I've been so overwhelmed with school and sports. There was a time when I enjoyed going to classes and going to practice every afternoon. Now, everything is piling up and wearing me down. Thankfully, I get to see you every day. You are truly the sunshine of my life. Thank you for making me laugh when I'm feeling down.
What is the context of the passage?
What is being compared in the metaphor?
What is the meaning of the metaphor?
Explanation:
Answer:
The solution code is written in Java.
System.out.println(numItems);
Explanation:
Java <em>println() </em>method can be used to display any string on the console terminal. We can use <em>println()</em> method to output the value held by variable <em>numItems.</em> The <em>numItems </em>is passed as the input parameter to <em>println()</em> and this will output the value of <em>numItems</em> to console terminal and at the same time the output with be ended with a newline automatically.
Answer:
938.7 milliseconds
Explanation:
Since the transmission rate is in bits, we will need to convert the packet size to Bits.
1 bytes = 8 bits
1 MiB = 2^20 bytes = 8 × 2^20 bits
5 MiB = 5 × 8 × 2^20 bits.
The formula for queueing delay of <em>n-th</em> packet is : (n - 1) × L/R
where L : packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate = 2.1 Gbps = 2.1 × 10^9 bits per second.
Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9
queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9
queueing delay for 48th packet = 0.938725181 seconds
queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds
