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Vitek1552 [10]
3 years ago
14

the attendance qt a seminar in a year 1 was 500 in year 2 the attendance changed by 100 what was the percent change in attendanc

e?
Physics
1 answer:
TiliK225 [7]3 years ago
5 0
20% change from year one to year two.
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The members of the track team will be able to run faster after they drink the new energy drink.” Explain why this statement is a
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A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

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If you increase the mass of an object and want to move an object a specific distance, what do you need to do
Alexxandr [17]

If you increase the mass of an object and want to move an object a specific distance, then you need to do extra work than the earlier

<h3>What is work done?</h3>

The total amount of energy transferred when a force is applied to move an object through some distance

Work Done = Force * Displacement

For example, let us suppose a force of 10 N is used to displace an object by a displacement of 5 m then the work done on the object can be calculated by the above-mentioned formula

work done = 10 N ×5 m

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Learn more about work done from here

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