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3 years ago

8

A resistance of 30 ohms is placed in a circuit with a 90 volt battery. What current flows in the circuit?

1 answer:

blagie [28]3 years ago

4 0

Answer:

3A

Explanation:

Using Ohms law U=I×R solve for I by I=U/R

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A plane surface 25 cm wide has its temperature maintained at 80°C. Atmospheric air at 25°C ows parallel to the surface

Gre4nikov [31]

Answer:

See the detailed answer in attached file.

Explanation :

3 0

3 years ago

Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.

levacccp [35]

Answer:

$\dot W_{out} = 3863.98\,kW$

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

$-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0$

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

$h_{in} = 3353.1\,\frac{kJ}{kg}$

Outlet (Liquid-Vapor Mixture)

$h_{out} = 890.1\,\frac{kJ}{kg}$

The power produced by the turbine is:

$\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})$

$\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )$

$\dot W_{out} = 3863.98\,kW$

8 0

3 years ago

A cold-rolled sheet metal that is 40 mm wide and has a thickness of 5.00 mm is going to be bent into a V shape with a 60° angle.

Artist 52 [7]

a= the force of gravity b= the amount of bicker to maple syrup ratio

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3 years ago

De acuerdo con el tabulador de puestos de una compañía, los salarios mensuales que obtienen los trabajadores son los que se mues

inessss [21]

Sorry don’t understand

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3 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 $\frac{m^{3} }{s}$ </h3>

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago