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2 years ago

25. How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?

Physics

1 answer:

Lina20 [59]2 years ago

4 0

Length=l=4m

$\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{l}{g}}$

$\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{4}{9.8}}$

$\\ \rm\rightarrowtail T=2\pi(0.63887)$

$\\ \rm\rightarrowtail T=1.27774\pi$

$\\ \rm\rightarrowtail T=4.012s$

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Answer:

Explanation:

Given

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where $\alpha$ is angular acceleration of pulley

$4.88=\alpha \times 5.2\times 10^{-2}$

$\alpha =93.84 rad/s^2$

And Tension in Rope

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T=8.364 N

and Tension will provide Torque

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$8.364\times 5.2\times 10^{-2}=I\times 93.84$

$I=0.463\times 10^{-2} kg-m^2$

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Thus mass is uniformly distributed or some more towards periphery of Pulley

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3 years ago

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a fast charged particle passes perpendicularly through a thin glass sheet of index of refraction 1.5. The particle emits light i

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<h3>Why is the refractive index important?</h3>

The higher the refractive index the slower the light travels, which causes a correspondingly increased change in the direction of the light within the material. What this means for lenses is that a higher refractive index material can bend the light more and allow the profile of the lens to be lower.

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option D pregnant women

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3 years ago

A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3 Pa s and

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The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

g = acceleration due to gravity = 9.8 m/s²,
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σ = density of water = 1.0 g/cm³ and
η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

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v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

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