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ankoles [38]
3 years ago
15

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres​

Physics
1 answer:
Scilla [17]3 years ago
7 0

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

       \text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}

Similarly,

      \text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}

We know the below formulas,

          \text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}

          \text {volume of wire}=\pi \times r^{2} \times l

When equating both the equations, we can find length of wire as below, where \pi=\frac{22}{7}

          \frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l

         \frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l

The \pi value gets cancelled as common on both sides, we get

           \frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l

The 10^{-6} value gets cancelled as common on both sides, we get

           l=4 \times 9=36 m

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3 years ago
A finite line of charge with linear charge density λ = 3.35 × 10-6 C/m, and length L = 0.808 m is located along the x axis (from
Andrews [41]

Answer: magnitude = 169.66N/C

direction = 8.6859°

Explanation:

Given from the question, we have that;

the Length of line of charge L = 0.808 m

Linear charge density λ = 3.35 × 10⁻⁶ C/m

charge q = -7.32 × 10⁻⁷ C

Coulombs force constant K = 1/(4π ε0) = 8.99 × 10⁹ N·m²/C².

NB. The picture uploaded gives a diagrammatic description of the problem.

From Pythagoras theorem we have,

tan Θ = 3.75 / (10.7-1.56)  

Θ = 22.3076 °

recall that the Electric field at point P due to the finite wire is;

È = Kλ (L / b(L+b)) Î .............. (1)

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹

where È rep the electric field.

from equation 1, we have that  

È = 8.99 × 10⁹ × 3.35 × 10⁻⁶ (0.808/ 10.7(10.7 – 0.808))

È = 0.229905 × 10³ N/C Î

recall also that the Electric field at point -P due to -q is;

È  = (8.99 × 10⁹ × 7.32 × 10⁻⁷) / ((3.75)² + (10.75-1.56)²) = 0.6742 × 10² N/C

where E = -E₁cosθÎ  + E₁sinθĴ

E = - 0.62446 ×10²Î   + 0.2562 ×10²Ĵ

The Resultant Electric charge Er is given as;

Er = 1.6771 ×10²Î + 0.2562 ×10²

Er =  [√(1.6771)² + (0.2562)² ] × 10² = 169.66 N/C

∴ Magnitude = 169.66 N/C

Having gotten the magnitude, let us find the direction;

⇒ Direction = tan Ф = 0.25621/1.6771 = 8.6859°

Direction = 8.6859°

cheers i hope this helps

7 0
3 years ago
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