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Monica [59]
3 years ago
8

A simple pendulum on the surface of Earth is found to undergo 15.0 complete small‑amplitude oscillations in 8.75 s. Find the pen

dulum's length.
Physics
1 answer:
FinnZ [79.3K]3 years ago
6 0

Answer:

Length of the Pendulum is 14.4 cm

Explanation:

Time period of simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g} } \\

time taken for one oscillation is

T=\frac{time taken}{oscillations} \\T=\frac{8.75}{15} \\T=0.583 s

Pendulum length on the earth surface is

L=\frac{T^2g}{4\pi^2 }\\L=\frac{0.583\times 9.8}{4\pi^2 } \\L=0.144 m

Length of the Pendulum is 14.4 cm

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In Trial III, a different, looser, spring is used; its force constant is 23.1 N/m. The suspended mass is the same as the one in
cricket20 [7]

Answer:

T=0.827s

Explanation:

The period of a spring can be calculated with the equation

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Gravity is affected by...
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A box has sides of 10 cm, 8.2 cm, and 3.5 cm. What is its volume?
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4 0
2 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

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0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

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6 0
3 years ago
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