what is the ampere unit of force and how is that unit derived?

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Answer 1

The ampere (/aempr/, US: /mpr/; symbol: A), also known as the amp, is the unit of electric current in the International System of Units (SI). One ampere is equal to one coulomb, or 6.2415090741018 electrons passing through a point in one second.

What is ampere?

The ampere, commonly abbreviated as "amp," is the base unit of electric flow in the International System of Units (SI). It was named after French mathematician and physicist André-Marie Ampère (1775-1836), widely regarded as the father of electrodynamics. The ampere is defined by the International System of Units in terms of other base units by estimating the electromagnetic power between electrical conductors that carry the electric flow. The previous CGS estimation framework employed two distinct definitions of current, one based on SI units and the other on electric charge as the base unit. The charge unit is calculated by measuring the power between two charged metal plates. The ampere was then defined as one coulomb of charge per second. SI defines the coulomb as the unit of charge.

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Related Questions

Suppose we have four jobs in a computer system, in the order JOB1, JOB2, JOB3 and JOB4. JOB1 requires 8 s of CPU time and 8 s of I/O time; JOB2 requires 4 s of CPU time and 14 s of disk time; JOB3 requires 6 s of CPU time; and, JOB4 requires 4 s of CPU time and 16 s of printer time. Define the following quantities for system utilization: • Turnaround time = actual time to complete a job • Throughput = average number of jobs completed per time period T • Processor utilization = percentage of time that the processor is active (not waiting) Compute these quantities (with illustrations if needed) in each of the following systems: a. A uniprogramming system, whereby each job executes to completion before the next job can start its execution. b. A multiprogramming system that follows a simple round-robin scheduling. Each process gets 2 s of CPU time turn-wise in a circular manner.

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Answer:

Explanation:

a. Uniprogramming System:

Turnaround Time: For JOB1, the turnaround time is 16 s (8 s CPU time + 8 s I/O time). For JOB2, the turnaround time is 18 s (4 s CPU time + 14 s disk time). For JOB3, the turnaround time is 6 s (6 s CPU time). For JOB4, the turnaround time is 20 s (4 s CPU time + 16 s printer time).

Throughput: In a uniprogramming system, the throughput is equal to 1 job per time period, as only one job can be processed at a time.

Processor Utilization: For JOB1, the processor utilization is 8/16 = 50%. For JOB2, the processor utilization is 4/18 = 22.2%. For JOB3, the processor utilization is 6/6 = 100%. For JOB4, the processor utilization is 4/20 = 20%.

b. Multiprogramming System:

Turnaround Time: For JOB1, the turnaround time is 24 s (8 s CPU time + 8 s I/O time + 8 s waiting). For JOB2, the turnaround time is 24 s (4 s CPU time + 14 s disk time + 6 s waiting). For JOB3, the turnaround time is 16 s (6 s CPU time + 10 s waiting). For JOB4, the turnaround time is 32 s (4 s CPU time + 16 s printer time + 12 s waiting).

Throughput: In a multiprogramming system, the throughput is equal to 4 jobs per time period, as all four jobs can be processed concurrently.

Processor Utilization: For JOB1, the processor utilization is 8/24 = 33.3%. For JOB2, the processor utilization is 4/24 = 16.7%. For JOB3, the processor utilization is 6/16 = 37.5%. For JOB4, the processor utilization is 4/32 = 12.5%.

Here are the turnaround time, throughput, and processor utilization for the four jobs in a uniprogramming and round-robin system:

Uniprogramming

Turnaround time:

JOB1: 16 s

JOB2: 22 s

JOB3: 12 s

JOB4: 20 s

Throughput: 1 job/16 s

Processor utilization: 50%

Round-Robin

Turnaround time:

JOB1: 18 s

JOB2: 14 s

JOB3: 10 s

JOB4: 18 s

Throughput: 2 jobs/16 s

Processor utilization: 62.5%

The difference in turnaround time between the two systems is due to the fact that in a uniprogramming system, the CPU is idle while a job is waiting for I/O. In a round-robin system, the CPU is always running, even if it is only running a short burst of code for another job.

The difference in throughput between the two systems is also due to the fact that in a round-robin system, the CPU is always running, so more jobs can be completed in a given amount of time.

The processor utilization is higher in a round-robin system because the CPU is not idle as often. However, the turnaround time is also higher because the jobs have to share the CPU.

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[Your Name]

[Your Address]

[City, State ZIP Code]

[Your Email]

[Date]

[Recipient Name]

[Recipient Position]

[Company Name]

[Address]

[City, State ZIP Code]

Dear [Recipient Name],

I am writing this letter to inform you that I have decided to step down from my position as [Your Position] at [Company Name]. My resignation is effective from [Date].

I have made this decision after much consideration and reflection on my personal and professional goals. While I have greatly enjoyed working at [Company Name] and appreciate the opportunities and support provided by the organization, I have decided that it is time for me to pursue other career opportunities.

I will ensure a smooth transition by completing my pending work and helping my replacement to settle into the role. I am willing to assist in any way possible to ensure that my duties and responsibilities are transferred smoothly to my successor.

I would like to take this opportunity to express my gratitude to [Recipient Name] and the entire management team for their support, guidance, and mentorship during my time at [Company Name]. I have learned a lot from my colleagues, and I am grateful for the professional growth opportunities provided to me.

Please let me know if there is any additional information you need from me or if I can assist in any way during this transition period.

Thank you for understanding my decision.

Sincerely,

[Your Name]

All of the following are common manual transmission problems EXCEPT: (A) bent shift mechanism. B) worn thrust washers. (C) clogged modulator valve. (D) damaged extension housing.

Answers

Option (C) clogged modulator valve is not a common manual transmission problem. This component is typically associated with automatic transmissions, not manual transmissions.

Explanation

Manual transmission problems can include a range of issues that can affect the performance of the transmission. Here short explanation about other option except C as common manual transmission problems:

Bent shift mechanism.
The shift mechanism in a manual transmission is responsible for moving the gears into the correct position for each gear change. If the shift mechanism is bent or damaged, it can cause difficulty in shifting gears or even prevent the transmission from shifting at all. This can happen due to wear and tear, damage from impact or other causes.Worn thrust washers.
Thrust washers are small components in the transmission that help to control the movement of the gears. If the thrust washers become worn, they can cause excessive play in the gears, leading to difficulty in shifting and even gear slippage. This can happen due to normal wear and tear or improper lubrication.Damaged extension housing.
The extension housing is the outermost component of the transmission on the backside that houses the output shaft. If it becomes damaged due to wear and tear, impact or other causes, it can cause leaks in the transmission or even prevent it from operating altogether. A damaged extension housing can cause the transmission to fail or operate poorly.

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how many grams of water can be heated from 10 degrees to 45 degrees using 20,500 joules of energy

Answers

Therefore, 20,500 Joules of energy can heat 1,104.3 grams of water from 10 degrees Celsius to 45 degrees Celsius.

Specific heat capacity

Calculate the energy (in Joules) required to heat 1 gram of water from 10 degrees Celsius to 45 degrees Celsius:

Q = m * c * ΔT

Q = 1 gram * 4.184 Joules/gram-degree Celsius * (45 - 10) degrees Celsius

Q = 18.47 Joules

Calculate the amount of water (in grams) that can be heated with 20,500 Joules of energy:

20,500 Joules / 18.47 Joules = 1,104.3 grams

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