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Alex777 [14]
2 years ago
8

The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Physics
1 answer:
VladimirAG [237]2 years ago
6 0

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

  • it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force  is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

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an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe
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Answer:

<em>The amount of electric charge transported = 0.192 C</em>

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

<em>Given:</em> I = 285 mA, t = 674 milliseconds.

<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>

<em>       (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s          </em>

Substituting these values into equation 1

Q = 0.285 × 0.674

<em>Q = 0.192 C</em>

<em>Therefore the amount of electric charge transported = 0.192 C</em>

<em></em>

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Explanation:

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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|

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|

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\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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