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pentagon [3]
2 years ago
12

1) Consider an electric power transmission line that carries a constant electric current of i = 500 A. The cylindrical copper ca

ble used to transmit this current has a diameter o = 2.00 cm and a length L = 150 km. If there are 8.43x10^28 free electrons per cubic meter (m^3 ) in the cable, calculate how long it would take for an electron to cross the entire length of the transmitter line.
Physics
1 answer:
JulsSmile [24]2 years ago
7 0

Answer:  

t = 1.27 x 10⁹ s  

Explanation:  

First, we will find the volume of the wire:

Volume = V = AL  

where,  

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²  

L = Length of wire = 150 km = 150000 m  

Therefore,    

V = 47.12 m³

 

Now, we will find the number of electrons in the wire:  

No. of electrons = n = (Electrons per unit Volume)(V)  

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)  

n = 3.97 x 10³⁰ electrons  

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

I =\frac{q}{t}  

where,  

t = time = ?  

I = current = 500 A  

q = total charge = (n)(chareg on one electron)  

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)  

q = 6.36 x 10¹¹ C  

500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}

Therefore,

<u>t = 1.27 x 10⁹ s</u>

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A group of 25 particles have the following speeds:
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Answer:

Part a: The average speed is 24.12 m/s

Part b: The rms speed is 25.55 m/s

Part c: The most probable speed is 17 m/s.

Explanation:

Part a

Average Speed

<em>Average speed is given as </em>

<em />v_{avg}=\frac{\sum}{n}

v_{avg}=\frac{(2 \times 11) +(7 \times 17)+(4 \times 19)+(3 \times 27) +(6 \times 32) +(1 \times 33)+(2 \times 40)  }{25}\\v_{avg}=\frac{22+119+76+81+192+33+80}{25}\\v_{avg}=\frac{603}{25}\\v_{avg}=24.12 m/s

So the average speed is 24.12 m/s

Part b

RMS Speed

v_{rms}=\frac{\sum v^2}{n}

v_{rms}=\sqrt{\frac{(2 \times 11^2) +(7 \times 17^2)+(4 \times 19^2)+(3 \times 27^2) +(6 \times 32^2) +(1 \times 33^2)+(2 \times 40^2)  }{25}} \\v_{rms}=\sqrt{\frac{242+2023+1444+2187+6144+1089+3200}{25}}\\\\v_{rms}=\sqrt{\frac{16329}{25}}\\v_{rms}=\sqrt{653.16}\\v_{rms}=25.55 m/s

So the rms speed is 25.55 m/s

Part c

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As 7 particles have speed of 17 m/s i.e. 7 is the highest frequency so 17 m/s is the most probable speed.

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Answer:

Gravity

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Normal force:

Cannot be included. Normal force is only applicable when object is on a surface, and it acts perpendicular to the surface. Since the ball is falling, there is no surface, and therefore no normal force. This question gives you unnecessary information, designed to trick you. Please remember when normal force is applicable.

Friction force:

Also only applicable when object is moving, and is on the table. Friction only applies when there is an applied force. There is no applied force when the ball is falling, so therefore no friction force.

Force of fall:

First of all, what is this? There is nothing called force of fall.

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A(n) 0.47 kg softball is pitched at a speed of 10 m/s. The batter hits it back directly at the pitcher at a speed of 29 m/s. The
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Answer:

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This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns

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