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2 years ago

In 1828, the diameter of the U.S. dime was changed to approximately 18 mm. What is this diameter when expressed in micrometers

Engineering

1 answer:

neonofarm [45]2 years ago

7 0

The diameter of an 18 mm dine when expressed in micrometers is; 18000 micrometers

<h3>Conversion of Units</h3>

We are given the diameter of dime at 18 mm.

Now, from conversion of units between mm(millimeter) and micrometer, we know that;

1 millimeter = 1000 micrometer.

Thus, by proportion, 18 millimeter will be;

18 millimeters × 1000 micrometers/1 millimeter

>> 18000 micrometers

In conclusion, the diameter of the 18mm fine in micrometers is; 18000 micrometers

Read more about conversion of units at; brainly.com/question/8421470

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___________ is NOT a common injury that an automotive tech may experience at work.

Degger [83]

Answer:The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

Explanation:

The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

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3 years ago

What type of circles have two or more circles with different center points?

Assoli18 [71]

Answer:

Concentric circles

Explanation:

Concentric circles are two or more circles which have the same center point. The region between two concentric circles is called an annulus.

6 0

3 years ago

Read 2 more answers

Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4

Ghella [55]

Answer:

0.19s

Explanation:

Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.

Queueing delay =(N-1) L /2R

where N = no of packet =93

L = size of packet = 4MB

R = bandwidth = 1.4Gbps = 1×10⁹ bps

4 MB = 4194304 Bytes

(93 - 1)4194304 / 2× 10⁹

queueing delay =192937984 ×10⁻⁹

=0.19s

5 0

3 years ago

1) Find the time in seconds to reach full charge in an RL circuit with L = 5 H and R = 100 ohms

pav-90 [236]

The time constant to reach full charge in an RL circuit is 0.05 ms.

Explanation:

To find the time constant,

The time constant for an RL circuit is defined by τ = L/R.

The given data is

L= 5 H

R= 100 ohms

by using the formula,

τ = L/R

= 5/100

= 0.05 ms

τ = 0.05 ms

Thus, the time constant to reach full charge in an RL circuit is 0.05 ms.

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3 years ago

A student lab group is brainstorming the design of an experiment that uses an ammeter (measures current) and different resistors

givi [52]

Answer:

Put a 10.0-ohm resistor in the circuit. Measure the current in the circuit. Replace the 10.0-ohm resistor with a 20.0-ohm resistor. Measure the new current. Continue replacing the resistor with a different resistor of known resistance. Measure the current for each resistor. Record all data.

Explanation:

The only design that has resistance varying with everything else remaining the same is the first design. That would be what you'd want to do if you're exploring the effect of resistance on current.

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3 years ago