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notka56 [123]
3 years ago
11

A wire of resistance R is cut into ten equal parts which are then connected in parallel. The equivalent resistance of the combin

ation is
Physics
1 answer:
Greeley [361]3 years ago
6 0

Answer:

<em>The equivalent resistance of the combination is R/100</em>

Explanation:

<u>Electric Resistance</u>

The electric resistance of a wire is directly proportional to its length. If a wire of resistance R is cut into 10 equal parts, then each part has a resistance of R/10.

Parallel connection of resistances: If R1, R2, R3,...., Rn are connected in parallel, the equivalent resistance is calculated as follows:

\displaystyle \frac{1}{R_e}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+...+\frac{1}{R_n}

If we have 10 wires of resistance R/10 each and connect them in parallel, the equivalent resistance is:

\displaystyle \frac{1}{R_e}=\frac{1}{R/10}+\frac{1}{R/10}+\frac{1}{R/10}...+\frac{1}{R/10}

This sum is repeated 10 times. Operating each term:

\displaystyle \frac{1}{R_e}=\frac{10}{R}+\frac{10}{R}+\frac{10}{R}+...+\frac{10}{R}

All the terms have the same denominator, thus:

\displaystyle \frac{1}{R_e}=10\frac{10}{R}=\frac{100}{R}

Taking the reciprocals:

R_e=R/100

The equivalent resistance of the combination is R/100

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Answer:

33,458.71 turns

Explanation:

Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter

μ₀ = Permeability of free space = 4 π × 10 ⁻⁷

Solution:

We have B = μ₀ × n × I

⇒ n = B/ (μ₀ × I)

n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)

n = 90,428.94 turn/m

No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37

= 33,458.71 turns

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The probability of an event A occurring is 0.73.
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A 5.0 kg box is sliding across a waxed floor by the application of a 15 N east force. If the force of friction is 2.5 N west, wh
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1) 12.5 N east

There are two forces acting on the box along the horizontal direction:

- The applied force of 15 N east, we indicate it with F

- The force of friction of 2.5 N west, we indicate it with F_f

Taking east as positive direction, we can write the two forces has

F=+15 N\\F_f = -2.5 N

Therefore, the net force on the box will be:

F_{net} = F + F_f = 15 + (-2.5) = +12.5 N

And the positive sign means the direction is east.

2) 2.5 m/s^2

We can solve this part by using Newton's second law:

F_{net}=ma

where

F_{net} is the net force on the box

m is its mass

a is the acceleration

For the box in this problem,

F_{net} = 12.5 m/s^2 (east)

m = 5.0 kg

Solving for a, we find the acceleration:

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6 0
3 years ago
Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same
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Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

\underset{V_{A}}{\rightarrow} = velocity of car A before collision = 0 i - V_{A} j

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Using conservation of momentum

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(1750) (0 i - V_{A} j) + (1450) (V_{B} i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) V_{B} i - (1750) V_{A} j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) V_{B} = 97600    and - (1750) V_{A} = - 60160

V_{B} = 67.3 km/h        and  V_{A} = 34.4 km/



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