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laila [671]
4 years ago
9

A 170 g air-track glider is attached to a spring. The glider is pushed in 11.2 cm against the spring, then released. A student w

ith a stopwatch finds that 14 oscillations take 11.0 s. What is the spring constant?
Physics
1 answer:
4vir4ik [10]4 years ago
7 0

Answer:k = 10.83 N/m²

Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below

ω = √k/m

But ω = 2πf, where f = frequency.

f = number of oscillations /time taken

Number of oscillations = 14, time taken = 11s

f = 14/11 = 1.27Hz.

ω = 2×22/7×1.27

ω = 7.98 rad/s.

By substituting this parameters into ω = √k/m

Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.

7.98 = √k/0.17

By squaring both sides

(7.98)² = k/ 0.17

k = (7.98)² × 0.17

k = 10.83 N/m²

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Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current
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force of attraction per unit length between the two current carrying wire is given by

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285 x 10⁻⁶ = 10⁻⁷ x \frac{2\times25.5\times I_2}{.3}

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To find point where magnetic field is zero due the these wires

The point will lie between the two wires  as current is in the same direction.

Let at y = y , the neutral point lies

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