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bagirrra123 [75]
3 years ago
7

What is the relationship between SI units and CGS unit of work ?

Physics
1 answer:
VashaNatasha [74]3 years ago
8 0
Work SI unit = joule = N*m= [kg]*[m/s^2] *[m] = kg * m^2/s^2

Work cgs unit = erg =  [g][cm/s^2][cm] = g*cm^2 / s^2

Then 1 kg * m^2 / s^2 * [1000 g/kg] * [100cm/m]^2 = 10,000,000 g*cm^2/s^2

The relation is 1 joule = 10,000,000 erg or 1 erg = 10^-7 joule
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Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2
aleksandr82 [10.1K]

Answer:

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

Explanation:

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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Answer:

The breaking in <em>molecular</em> bonds in food releases energy for your body to use.

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