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Anna007 [38]
3 years ago
9

City A and city B are located at the same latitude and elevation. City A is a coastal city, and city B is located far inland. A

cold surface current flows by the coastline where city A is located. Which statement is most likely true for these cities?
Choose the correct answer.


During the fall, city B experiences more hurricanes than city A experiences.

During the summer, city B experiences cooler weather than city A experiences.

During the winter, city B experiences warmer weather than city A experiences.

During the spring, city B experiences more thunderstorms than city A experiences.
Engineering
2 answers:
jenyasd209 [6]3 years ago
8 0

Answer:

Option 4

Explanation:

IceJOKER [234]3 years ago
6 0
The third option is correct
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Complete the following sentence. The skills and content of several subject areas were combined to form a new field known as a me
Svetlanka [38]

Answer:

The answer is "discipline" , I was doing the same thing a minute ago. I hope this anwser is helpful! ^^

Explanation:

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3 years ago
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sa kasalukuyang panahon posible pakayang mag karoon ng panibagong kilusang propaganda at lilusang katipunan oo?at bakit​
gladu [14]

Answer:

?

Explanation:

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Something that pollutes a system is called a(n) ??? .
frutty [35]

Answer:

Trash pollution

correct me if wrong

Explanation:

7 0
3 years ago
What is the non-linearity error, as a percentage of full range, produced when a 1K Ω potentiometer has a load of 10KΩ & is a
Darina [25.2K]

Answer:

2.28%

Explanation:

Being at one third of its maximum range a potentiometer should output V0/3.

However if this 1kΩ potentiometer has a 10kΩ load:

(1) I1 = I2 + I3

(2) V0 = I1 * 2/3*Rp + I3 * 1/*Rp

(3) Vp = I2 * Rl

(4) Vp = I3 * 1/3 * Rp

Where

I1: current entering the potentiometer

I2: current going to the load

I3: current going to the other leg of the potentiometer

V0: supply voltage

Vp: output voltage of the potentiometer

Rp: total resistance of the potentiometer

Rl: load resistance

First we determine the intensity of I3 in function of supply power

I3 = 3 * Vp / Rp  = 3 * Vp / 1000 = 0.003*Vp

Then the load current

I2 = Vp / Rl  = Vp / 10000 = 0.0001*Vp

With these we determine I1

I1 = 0.003*Vp + 0.0001*Vp = 0.0031*Vp

Then

V0 = 0.0031 * Vp * 2/3*Rp + 0.003*Vp * 1/3*Rp

V0 = 0.00207 * Vp * Rp + 0.001 * Vp * Rp

V0 = 0.00307 * Vp * 1000

V0 = 3.07 * Vp

Vp = V0 / 3.07

Vp = 0.3257 * V0

Now the percentage error is:

(100 * (0.3333 - 0.3257)) / 0.3333 = 2.28 %

7 0
4 years ago
Given the circuit at the right in which the following values are used: R1 = 20 kΩ, R2 = 12 kΩ, C = 10 µ F, and ε = 25 V. You clo
agasfer [191]

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

8 0
3 years ago
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