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3 years ago

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When you push a child on a swing, your action is most effective when your pushes are timed to coincide with the natural frequenc

y of the motion. You are swinging a 30 kg child on a swing suspended from 5 meter cables.Estimate the optimum time interval between your pushes. Would your answer be different if the child had a different mass?

1 answer:

OleMash [197]3 years ago

3 0

Answer:

$T = 4.48 s$

we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different

Explanation:

Natural frequency of a simple pendulum of L length is given as

$f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}$

so the time period of the oscillation is given as

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

so we will have

$L = 5 m$

$T = 2\pi\sqrt{\frac{5}{9.81}}$

$T = 4.48 s$

also from above formula we can see that this time period is independent of the mass of the child so answer would be same if the child mass is different

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Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

∫ E ds represents total electric flux passing through the closed surface due to charge q enclosed in the surface .

This also represents total flux coming out of the charge q on all sides .

This is equal to q / ε where ε is a constant called permittivity which depends upon the medium enclosing the charge . For air , its value is 8.85 x 10⁻¹² .

If charge remains the same but radius of the sphere enclosing the charge is doubled , the flux coming out of charge will remain the same .

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A force of 100. newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the time it takes to do

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<h3>It takes 60 seconds to do the work</h3>

<em><u>Solution:</u></em>

Given that,

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Power = 25 watts

To find: time it takes to do the work

<em><u>Find the work done:</u></em>

$work = force \times displacement\\\\work = 100\ newtons \times 15\ meters\\\\work = 1500\ joule$

<em><u>Find the time taken</u></em>

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3 years ago

Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa

anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are $146.03\times10^{3}\ N/C$ and 75.36°.

Explanation:

Given that,

First charge $q_{1}= 2.9\mu C$

Second charge $q_{2}= 2.9\mu C$

Distance between two corners r= 50 cm

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Using formula of electric field

$E_{1}=\dfrac{kq}{r'^2}$

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For E₂,

Using formula of electric field

$E_{1}=\dfrac{kq}{r^2}$

Put the value into the formula

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$E_{x}=52.2\times10^{3}\times\cos45$

$E_{x}=36910.97=36.9\times10^{3}\ N/C$

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$E_{y}=E_{2}+E_{1}\sin\theta$

$E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45$

$E_{y}=141310.97=141.3\times10^{3}\ N/C$

We need to calculate the net electric field

$E_{net}=\sqrt{E_{x}^2+E_{y}^2}$

Put the value into the formula

$E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}$

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Using formula of direction

$\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}$

$\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})$

$\theta=75.36^{\circ}$

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2 years ago

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1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

$n=\frac{c}{v}$

where

c is the speed of light in a vacuum

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Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

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where

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