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Elanso [62]
3 years ago
11

"A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. When the system reaches it

s steady state, what is the voltage across the resistor and the capacitor?
a. The voltage across both the resistor and the capacitor is equal to one-half of the terminal voltage of the battery.
b. The voltage across the resistor is equal to the terminal voltage of the battery, and the voltage across the capacitor is zero.
c. The voltage across the resistor is zero, and the voltage across the capacitor is equal to the terminal voltage of the battery.
d. The voltage across both the resistor and the capacitor is equal to the terminal voltage of the battery.
e. The voltage across both the resistor and the capacitor is zero"
Physics
1 answer:
sashaice [31]3 years ago
6 0

Answer:

b. The voltage across the resistor is equal to the terminal voltage of the battery, and the voltage across the capacitor is zero.

Explanation:

<em>When the system reaches its steady state the current tends to zero, then</em>

<em>in the resistor</em>

<em>V=IR</em>

<em>I=0, then </em><em>V=0</em>

<em>the capacitor charged is like an interruption in the circuit. </em><em>If we mesure voltage on both sides of the battery and if we mesure both sides of the capacitor, we will get the same value</em><em>, as its not any component between them in one side, and, in the other, the only component is the resistor that, without any courrent doesn't discipates any energy.</em>

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Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

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(d)   Δy = 5.11*10^-3m

Explanation:

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v=\sqrt{2gh}        (1)

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(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

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You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

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(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

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THe compression of the ball when it strikes the floor is 5.11*10^-3m

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