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daser333 [38]
3 years ago
6

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

w many time constants (a decimal number) must elapse before a capacitor in a series RC circuit is charged to 39.0% of its equilibrium charge
Physics
1 answer:
Evgesh-ka [11]3 years ago
7 0

Answer: t = 0.492τ

Explanation: In a circuit where there is a resistor and a capacitor, the equation for a charging capacitor is given by:

q = q_0(1 - e^{\frac{t}{RC} })

where:

q_0 is the equilibrium charge

q is the charge at time t

RC is time constant also called τ (tau)

For this problem, the circuit is charged to 39%, which means: q = 0.39 q_0

0.39q_0 = q_0 (1 - e^\frac{-t}{RC} )

0.39 = 1 - e^\frac{-t}{RC}

e^\frac{-t}{RC} = 0.61

\frac{-t}{RC} = ln(0.61)

-t = ln(0.61)τ

t = 0.492τ

For the condition to be met it is needed 0.492 time constants must elapse.

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Please help The position of masses 4kg, 6kg, 7kg, 10kg, 2kg, and 12kg are (-1,1), (4,2), (-3,-2), (5,-4), (-2,4) and (3,-5) resp
Sophie [7]

Answer:

(1.9756, -2.1951)

Explanation:

The center of mass equation is: x_{cm} = \frac{m_{1}x_{1}   + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}}, where m represents the masses and x represents the position.

In order to find the coordinates of the center of mass, we need to use this equation for both the x-values and the y-values.

<u>x-values:</u>

<u />x_{cm} = \frac{m_{1}x_{1}   + m_{2}x_{2} + m_{3}x_{3} + m_{4}x_{4} + m_{5}x_{5} + m_{6}x_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}} = \frac{4(-1)+6(4)+7(-3)+10(5)+2(-2)+12(3)}{4+6+7+10+2+12} = \frac{(-4)+(24)+(-21)+(50)+(-4)+(36)}{41} = \frac{81}{41} = 1.9756

<u>y-values:</u>

<u />y_{cm} = \frac{m_{1}y_{1}   + m_{2}y_{2} + m_{3}y_{3} + m_{4}y_{4} + m_{5}y_{5} + m_{6}y_{6}}{m_{1} + m_{2} + m_{3} + m_{4} + m_{5} + m_{6}} = \frac{4(1)+6(2)+7(-2)+10(-4)+2(4)+12(-5)}{4+6+7+10+2+12} = \frac{(4)+(12)+(-14)+(-40)+(8)+(-60)}{41} = \frac{-90}{41} = -2.1951

<u>center of mass:</u>

(1.9756, -2.1951)

6 0
3 years ago
Figure one, voltmeters
slava [35]

Answer:

(i) Half

(ii) 3 V

(iii) V₁

Explanation:

(i) The given parameters are;

The circuits have identical resistances

The number of resistors in circuit 1 = 1 resistor

The number of resistors in circuit 2 = 2 resistors

Let 'R' represent the value of each resistor, we have;

The total resistance of circuit 1 = R Ohm

The total resistance of circuit 3 = 2·R Ohm

∴ The total resistance of circuit 1  = (1/2) × The total resistance of circuit 3

∴ The resistance of circuit 1 is <u>half</u> the resistance of circuit 3

(ii) The potential difference of each cell, V = 1.5 volts

The number of cells in circuit 2 = 2 cells

The total potential difference of the cells of circuit 2 = 2 × 1.5 volts = 3 × volts = 3 V.

The voltmeter reading = The potential difference across the cell or cells it is applied

∴ The voltmeter reading on voltmeter, V₂, applied across the cells of circuit 2 = 3 V

(iii) The voltmeter reading V₁ = 1.5 V

The voltmeter reading V₂ = 3 V

The voltmeter reading V₃ = 4.5/(2·R) × R = 2.25 V

Therefore, the voltmeter reading with the smallest volt, is V₁ = 1.5 V

6 0
3 years ago
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