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garri49 [273]
2 years ago
12

What must be the length of a simple pendulum if its oscillation frequency is to be equal to that of an air-track glider of mass

0.250 kg attached to a spring of force constant 9.75 N/m
Physics
1 answer:
Anvisha [2.4K]2 years ago
8 0

Answer:

the length of the simple pendulum is 0.25 m.

Explanation:

Given;

mass of the air-track glider, m = 0.25 kg

spring constant, k = 9.75 N/m

let the length of the simple pendulum = L

let the frequency of the air-track glider which is equal to frequency of simple pendulum = F

The oscillation frequency of air-track glider is calculated as;

F = \frac{1}{2\pi } \sqrt{\frac{k}{m} } \\\\F = \frac{1}{2\pi } \sqrt{\frac{9.75}{0.25} } \\\\F = 0.994 \ Hz

The frequency of the simple pendulum is given as;

F = \frac{1}{2\pi} \sqrt{\frac{g}{l} } \\\\2\pi(F) = \sqrt{\frac{g}{l} } \\\\2\pi (0.994) = \sqrt{\frac{9.8}{l} } \\\\6.2455 = \sqrt{\frac{9.8}{l} } \\\\(6.2455)2 = \frac{9.8}{l} \\\\39.006 = \frac{9.8}{l} \\\\l = \frac{9.8}{39.006} \\\\l = 0.25 \ m

Thus, the length of the simple pendulum is 0.25 m.

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A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands
pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

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Distance of object = 12 cm

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Using formula of lens

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\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}

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The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

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Using formula of lens

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A body, with a volume of 2 m3, weighs 40 kN. Determine its weight when
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8.8 kN

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W = 6642 J

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So, the work done is 6642 J.

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2 years ago
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