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Olin [163]
3 years ago
7

A resistor R and a capacitor C are connected in series to a battery of terminal voltage V0. Which of the following equations rel

ating the current I in the circuit and the charge Q on the capacitor describes this circuit?
(A) V0 + QC – I2R = 0
(B) V0 – Q/C – IR = 0
(C) V02 – Q2/2C – I2R = 0
(D) V0 – CI – I2R = 0
(E) Q/C – IR = 0
Physics
1 answer:
Lesechka [4]3 years ago
5 0
Answer-C VO2-Q2/2C-I2R=0
Q1/2C currents upon the charge I2R on the capacitor. Voltage V02 goes best
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Charging an object without touching it is called?
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<span>Charging by induction</span>
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Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
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Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

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4 years ago
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