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Nostrana [21]
3 years ago
5

Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is

4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?
Physics
1 answer:
Alex777 [14]3 years ago
6 0

Answer:

<em>10.75 A</em>

<em>The current is in opposite direction since it causes a repulsion force between the wires</em>

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

F/l = \frac{u_{0}I_{1} I_{2}  }{2\pi r }

where F/l is the force per unit length on the wires

u_{0} = permeability of vacuum = 4π × 10^−7 T-m/A

I_{1} = current on the first wire = 0.520 A

I_{2} = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = \frac{4\pi *10^{-7}*0.520*I_{2}  }{2\pi *0.026} =  \frac{ 2*10^{-7}*0.520*I_{2}  }{0.026}

4.30×10^−5 = 4 x 10^-6 I_{2}

I_{2} = (4.30×10^-5)/(4 x 10^-6) = <em>10.75 A</em>

<em>The current is in opposite direction since it causes a repulsion force between the wires.</em>

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3 years ago
The Shinkansen (bullet train in Japan) makes a trip from Tokyo Station to Kyoto station in 2 hours and 14 min. The distance trav
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Answer:

v =  57.2 m/s

Explanation:

The average velocity of the train can be defined as the total distance covered by the train divided by the time taken by the train to cover that distance. Therefore, we will use the following formula to find the average velocity of the train:

v = s/t

where,

s = distance covered = 460 km = (460 km)(1000 m/1 km) = 4.6 x 10⁵ m

t = time taken to cover the distance = 2 h 14 min

Now, we convert it into minutes:

t = (2 h)(60 min/1 h) + 14 min

t = 120 min + 14 min = (134 min)(60 s/1 min)

t = 8040 s

Therefore, the value of velocity will be:

v = (4.6 x 10⁵ m)/8040 s

<u>v =  57.2 m/s</u>

7 0
3 years ago
A circuit consists of a series combination of 6.50 −kΩ and 4.50 −kΩ resistors connected across a 50.0-V battery having negligibl
Vlada [557]

Answer:

Part A: 16.1 V

Part B: 20.5 V

Part C: 21.5%

Explanation:

The voltmeter is in parallel with the 4.5-kΩ resistor and the combination is in series with the 6.5-kΩ resistor. The equivalent resistance of the parallel combination is given as

\dfrac{1}{R_E}=\dfrac{1}{4.50}+\dfrac{1}{10.0}

R_E=\dfrac{4.50\times10.0}{4.50+10.0} = 3.10

Part A

The voltmeter reading is the potential difference across the parallel combination. This is found by using the voltage-divider rule.

V_1 = \dfrac{3.10}{3.10+6.50}\times50.0 = \dfrac{3.10}{9.60}\times50.0 = 16.1 \text{ V}

Part B

Without the voltmeter, the potential difference across the 4.5-kΩ resistor is found using the same rule as above:

V_2 = \dfrac{4.50}{4.50+6.50}\times50.0 = \dfrac{4.50}{11.0}\times50.0 = 20.5 \text{ V}

Part C

The error in % is given by

\dfrac{20.5-16.1}{20.5}\times100\% = \dfrac{4.4}{20.5}\times100\% = 21.5\%

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3 years ago
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Explanation:

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Example 3 :
kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d =  diameter = 120mm = 0. 12m

V =  velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × \frac{0. 06}{0. 12* 1* 0. 9}

friction factor = 0. 52 × \frac{0. 06}{0. 108}

friction factor = 0. 52 × 0. 55

friction factor = 0. 289

b. When V = 3mls

Friction factor = 0. 52 × \frac{0. 06}{0. 12 * 3* 0. 9}

Friction factor = 0. 52 × \frac{0. 06}{0. 324}

Friction factor = 0. 52 × 0. 185

Friction factor = 0.096

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × \frac{1}{2*6. 6743 * 10^-11} × \frac{1^2}{0. 120} × \frac{1}{100}

Head loss =  1. 80 × 10^8

Head loss When V = 3m/s

Head loss = 0. 096 × \frac{1}{1. 334 *10^-10} × \frac{3^2}{0. 120} × \frac{1}{100}

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss  when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

Learn more about friction here:

brainly.com/question/24338873

#SPJ1

4 0
1 year ago
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