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jonny [76]
3 years ago
9

A certain machine changes a large input force into a smaller output force. how will the machine affect the distance over which t

he force is applied? explain.
Physics
1 answer:
Molodets [167]3 years ago
6 0

A machine that lowers the input force is a step down machine. If the force is lowered, the distance at which the the force is applied will be lowered as well. Assuming that the work done by the force does not stay constant and is affected by force, the distance will be reduced.
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Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
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Answer:

The curl is 0 \hat x -z^2 \hat y -4xy \hat z

Explanation:

Given the vector function

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\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|

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\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|

So we will get

\nabla  \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z

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\nabla  \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z

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