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3 years ago

What are the rules for setting up an integral of rotation?

1 answer:

3 0

Setting up an integral of rotation is used as a method of of calculating the volume of a 3D object formed by a rotated area of a 2D space. Finding the volume is similar to finding the area, but there is one additional component of rotating the area around a line of symmetry.

<span>First the solid of revolution should be defined. The general function is y=f(x), on an interval [a,b].</span>

Then the curve is rotated about a given axis to get the surface of the solid of revolution. That is the integral of the function.

<span>It all depends of the function f(x), which must be known in order to calculate the integral.</span>

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4 (JAMB) The diagram below shows a light see-saw, which is balanced horizontally by the weights W₁, W2, W3, W4 in the positions

damaskus [11]

The equation that represents the principle of the lever balance is:

W₁ + W₂ = W3 + W4; option A.

<h3>What is the principle of moments?</h3>

The principle of moments states when a body is in equilibrium, the sum of the clockwise moment about a point equals the sum of anticlockwise moment about that point.

A see-saw represents a balanced system of moments.

The sum of clockwise moment = The sum of anticlockwise moments.

Assuming W1 and W2 are clockwise moments and W3 and W4 are anticlockwise moments.

The equation will b: W₁ + W₂ = W3 + W4

In conclusion, a balanced see-saw illustrates the principle of the lever balance.

Learn more about principle of moments at: brainly.com/question/20519177

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2 years ago

Consider an insulating sphere of radius 6 cm surrounded by a conducting sphere of inner radius 18 cm and outer radius 26 cm. Fur

dimulka [17.4K]

Answer:

$-1.7908787542\times 10^{-9}\ C$

$3.4260289211\times 10^{-9}\ C$

$1.7908787542\times 10^{-9}\ C$

$1.6351501669\times 10^{-9}\ C$

Explanation:

r = Radius

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Electric field is given by

$E=-\dfrac{kq}{r^2}\\\Rightarrow q=-\dfrac{Er^2}{k}\\\Rightarrow q=-\dfrac{1610\times 0.1^2}{8.99\times 10^9}\\\Rightarrow q=-1.7908787542\times 10^{-9}\ C$

The charge is $-1.7908787542\times 10^{-9}\ C$

$Q+q=\dfrac{Er^2}{k}\\\Rightarrow Q=\dfrac{Er^2}{k}-q\\\Rightarrow Q=\dfrac{120\times 0.35^2}{8.99\times 10^9}-(-1.7908787542\times 10^{-9})\\\Rightarrow Q=3.4260289211\times 10^{-9}\ C$

The charge is $3.4260289211\times 10^{-9}\ C$

The charge inside will have the polarity changed

$q=+1.7908787542\times 10^{-9}\ C$

Outside the charge will be

$3.4260289211\times 10^{-9}-1.7908787542\times 10^{-9}\\ =1.6351501669\times 10^{-9}\ C$

3 0

3 years ago