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Leona [35]
3 years ago
10

An uncharged capacitor is connected to a resistor and a battery. Choose what happens to current, potential difference and charge

right after the circuit is closed. An uncharged capacitor is connected to a resistor and a battery. Choose what happens to current, potential difference and charge right after the circuit is closed.
Potential difference across the capacitor starts high and then drops exponentially.
Current through the circuit starts with zero and then increases gradually to a maximum value.
Charge on the plates of the capacitor decreases with time.
Charge on the plates of the capacitor increases with time.
Charge on the plates of the capacitor doesn't change with time.
Potential difference across the capacitor starts with zero and then increases gradually to a maximum value.
Current through the circuit starts high and then drops exponentially.
Engineering
1 answer:
Ilya [14]3 years ago
6 0

Answer:

  • The charge on the plates will increase with time
  • The potential difference across the capacitor starts with zero and then increases gradually to a maximum value
  • The current through the circuit starts high and then drops exponentially

Explanation:

<u>Case : An uncharged capacitor is connected to a resistor and a battery in a closed circuit.</u>

  • The charge on the plates will increase with time

applying this equation : Q = Q_{0}  [ 1 - e^{\frac{-t}{RC} } ]  as the value of (t) increases the value of Q increases i.e. charge on the plates

  • The potential difference across the capacitor starts with zero and then increases gradually to a maximum value

applying this equation : V = V_{0}  [ 1 - e^{\frac{-t}{RC} } ]

  • The current through the circuit starts high and then drops exponentially

current : I = I_{0}  e^{\frac{-t}{RC} }

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4 years ago
Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the
myrzilka [38]

Answer:

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}

The head associated with the pump is cleared:

h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})

Inlet and outlet velocities are found:

v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }

v_{1} \approx 4.278\,\frac{m}{s}

v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }

v_{2} \approx 11.573\,\frac{m}{s}

Now, the head associated with the pump is finally computed:

h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m

h_{pump} = 7.705\,m

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\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}

\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)

\dot W_{pump} = 16264.922\,W\,(16.265\,kW)

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Nonamiya [84]
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What does WCS stand for? A. Western CAD System B. Worldwide Coordinate Sectors C. World Coordinate System D. Wrong CAD Settings
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Answer:

The correct answer is C. World Coordinate System

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7 0
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Answer:

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Explanation:

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We know that discharge through venturi meter

 Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

h=x(\dfrac{S_m}{S_w}-1)

S_m=13.6 for Hg and S_w=1 for water.

h=0.235(\dfrac{13.6}{1}-1)

h=2.961 m

Now by putting the all value in

Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt{A_1^2-A_2^2}}

0.004=0.8\times \dfrac{3.09\times 10^{-3} A_2\sqrt{2\times 9.81\times 2.961}}{\sqrt{(3.09\times 10^{-3})^2-A_2^2}}

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So throat diameter d_2=28.60 mm

     

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