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matrenka [14]
3 years ago
9

A switch that connects a battery to a 30μf capacitor is closed. several seconds later you find that the capacitor plates are cha

rged to ±30μc. part a what is the emf of the battery?
Physics
1 answer:
Elanso [62]3 years ago
5 0
You can find the emf of the battery by using:

Q = CV.

Q = Charge in capacitor plates = 30μC
C = Capacitance of the capacitor = 30μF

V = Q/C = 30/30 = 1V

Ans: The emf of the battery = 1V.
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Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo
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Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

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Therefore,

M₀ = (i + j + k) x (1i + 0j +  5k)

M₀ = \left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

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Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0
3 years ago
A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t
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Answer:

v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

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Second stone:

y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}

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PtichkaEL [24]
Hiii !!!
I am sending the soluction !!!

If you have any question, let me now =)

Jorge:)

4 0
3 years ago
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