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ira [324]
3 years ago
12

Solve at least one question (30 points)

Physics
1 answer:
EastWind [94]3 years ago
4 0

Answer:

(a) the particle position = 135 m

(b) the velocity of the particle = 44 m/s

(c) the acceleration of the particle = 50 m/s²

Explanation:

Solution to Question 2.

Given;

velocity of a particle, v = 2 - 4t + 2t³

initial position at t = 0, s₀ = 3 m

(a) the particle position at t = 3 s

s = vt

s = (2 - 4t + 2t³)t

s = 2t - 4t² + 2t⁴

s = s₀ + s₃

s = s₀ + 2(3) - 4(3²) + 2(3⁴)

s = s₀ + 6 - 36 + 162

s = s₀ + 132 m

s = 3m + 132 m

s = 135 m

(b) the velocity of the particle at t = 3 s

v = 2 - 4t + 2t³ = 2 - 4(3) + 2(3)³

v = 44 m/s

(c) the acceleration of the particle at t = 3s

v = 2 - 4t + 2t³

a = \frac{dv}{dt} \\\\a = \frac{d}{dt} (2 -4t + 2t^3)\\\\a = -4 + 6t^2

a = -4 + 6(3)²

a = 50 m/s²

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<u>Given:</u>

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\oint \vec B \cdot d\vec l=\mu_o I.

In case of a circular loop, the directions of magnetic field and the line element d\vec l, both are along the tangent of the loop at that point, therefore, \vec B\cdot d\vec l = B\ dl.

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Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, r_1 = 3.510\ m.

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The magnetic field at the given point due to this wire is given by:

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B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.

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