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emmainna [20.7K]
4 years ago
14

What potential difference is required to cause 4.00 a to flow through a resistance of 330 ω?

Physics
1 answer:
Alisiya [41]4 years ago
7 0
We can solve the problem by using Ohm's law, which states that an Ohmic conductor the following relationship holds:
\Delta V = I R
where
\Delta V is the potential difference applied to the resistor
I is the current flowing through it
R is the resistance

In our problem, I=4.00 A and R=330 \omega, so the potential difference is
\Delta V = IR=(4.00 A)(330 \omega)=1320 V
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A student leaving school travels 100 m East before he realizes he left his textbook in his locker. He heads back towards school,
andrew11 [14]

-- The overall <em>distance</em> he travels is (100m + 30m + 70m) = <em>200 meters</em>.

-- His <em>displacement </em>when he arrives at his front door is

D = (100m East) + (30m West) + (70m East)

D = (100m + 70m)East + (30m)West

D = (170m East) + (30m West)

<em>D = 140 meters East </em>

It's interesting to notice that his displacement is 60 meters shorter than the distance he walked.  

That's because there's a stretch of 30 meters somewhere in the middle that he actually covered <em>three times</em>.

Two of those times added to the distance his shoes covered (2x30m=60m), but they cancelled out of the displacement.

His front door is 140 meters East of school.  He walked 60m farther than that, going back and forth over the 30m piece.

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4 years ago
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3 years ago
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Jack and Nas decided to have a race from their houses to the park to know who runs the fastest. Jack whose house is 15 kilometer
Anettt [7]

Answer:

the Jack speed is greater than the Nas speed  

   v₁ = 2.08 m / s >  v2 = 1.85 m / s

Explanation:

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          v = x / t

Jack.

It tells us that it travels x = 15 km in a time of t = 2 h

let's reduce the magnitudes to the SI system

     x = 15 km (1000 m / 1 km) = 15 10³ m

     t = 2 h (3600 s / 1 h) = 7200 s

let's calculate

     v₁ = 15 10³/7200

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Nas travels a distance of x = 10 km in a time of t = 1.5 h

      x = 10 km = 10 10³ m

      t = 1.5 h (3600s / 1h) = 5400 s

let's calculate the speed

     v2 = 10 10³/5400

      v2 = 1.85 m / s

From these results we can see that the Jack speed is greater than the Nas speed

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3 years ago
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