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Karo-lina-s [1.5K]
4 years ago
13

Compare the work done by the electrical field when the particle travels

Physics
1 answer:
Svetach [21]4 years ago
4 0
Show transcribed image text. Compare the work done by the electric field when the particle travels<span> from point W to point X to that </span>done<span> when the </span>particle travels<span> from point Z to point Y. ... Is the </span>work done<span> on the </span>particle<span> by the </span>electric field<span> positive, negative, or zero? Explain using force and displacement vectors.</span>
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A statue is made of bronze, which is an alloy of about 80% copper and 20% tin. Which metal is in the higher concentration in thi
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Copper is the higher concentrated metal in this case.
7 0
3 years ago
Help with 1 2 and 3 please
geniusboy [140]

1. Amperes, is the SI unit (also a fundamental unit) responsible for current.

2. I = \frac{q}{t} Δq over Δt technically

Rearrange for Δq

I x Δt = Δq

1.5mA x 5 = Δq

Δq = 0.0075

Divide this by the fundamental charge "e"

Electrons: 0.0075 / 1.60 x 10^-19

Electrons: 4.6875 x 10^16 or 4.7 x 10^16

3.  So we know that the end resistances will be equal so:

ρ = RA/L

ρL = RA

ρL/A = R

Now we can set up two equations one for the resistance of the aluminum bar and one for the copper: Where 1 represents aluminum and 2 represents copper

\frac{p1L1}{A1}  = \frac{p2L2}{A2}\\

We are looking for L2 so we can isolate using algebra to get:

\frac{A2(\frac{P1L1}{A1}) }{P2} = L2

If you fill in those values you get 0.0205

or 2.05 cm



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3 years ago
The product of an object’s mass and velocity
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momentum is the product of mass and velocity

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3 years ago
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A guitar string is plucked. What best describes the formation and transmission of the sound wave?
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The stings vibrates around the air which makes a longitudinal wave

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An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
3 years ago
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