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zepelin [54]
2 years ago
10

Three resistors of 100 W, 3900 W, and 1000 W are connected in series across a 200-V battery. What is the voltage drop across the

resistor of value 1000 W?
Physics
1 answer:
Gala2k [10]2 years ago
7 0

Answer:

40 V

Explanation:

 I will assume that  the resistors are

           100 and 3900 and 1000 OHMS    <=====(NOT W)

In series , the resistances add together 100 + 3900 + 1000 = 5000 ohms total

V = IR

I = V / R      so the total current will be   200 v / 5000 ohms = .04 amps

                   this is the current through all of the resistors

      so for the   1000 ohm resistor     V = IR     .04 (1000) = 40 V

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A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then t
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The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

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Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure (P_{atm}), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:

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

Where:

P_{1}, P_{2} - Water total pressures inside the tank and at ground level, measured in pascals.

\rho - Water density, measured in kilograms per cubic meter.

g - Gravitational acceleration, measured in meters per square second.

v_{1}, v_{2} - Water speeds inside the tank and at the ground level, measured in meters per second.

z_{1}, z_{2} - Heights of the tank and ground level, measured in meters.

Given that P_{1} = P_{2} = P_{atm}, \rho = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}}, v_{1} = 0\,\frac{m}{s}, z_{1} = 6.9\,m and z_{2} = 4.9\,m, the expression is reduced to this:

\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)

And final speed is now calculated after clearing it:

v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}

v_{2} \approx 6.263\,\frac{m}{s}

The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.

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4 years ago
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