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Alex777 [14]
3 years ago
8

Casey is playing with her yo-yo. She performs an "around the world" trick, in which the yo-yo travels up and around in a circle

before returning to her hand. At which point in its circular path would the yo-yo have the least potential energy?
Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

It has the least potential energy at the bottom of its circular path.

Explanation:

It has the least potential energy at the bottom of its circular path.

Remember the equation

U = m*g*h

where U is the potential energy

m is the mass of the yo-yo

h is the height

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Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

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

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Given the vector function

\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z

We can calculate the curl using the definition

\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|

Thus for the exercise we will have

\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

Working with the partial derivatives we get the curl

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

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Average velocity is 1..2 mi/min east

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