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3 years ago

What is the power of a parallel circuit with a resistance of 1,000 omh and current of 0.03a

Engineering

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer: P = I2R = 0.032 x 1000 =0.9 W

Explanation: The power will be the product of the square of the current and

the resistance of the load. The fact that the circuit is a parallel circuit is irrelevant to this question.

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A plate and frame heat exchanger has 15 plates made of stainless steel that are 1 m tall. The plates are 1 mm thick and 0.6 m wi

hodyreva [135]

Answer:

14.506°C

Explanation:

Given data :

flow rate of water been cooled = 0.011 m^3/s

inlet temp = 30°C + 273 = 303 k

cooling medium temperature = 6°C + 273 = 279 k

flow rate of cooling medium = 0.02 m^3/s

Determine the outlet temperature

we can determine the outlet temperature by applying the relation below

Heat gained by cooling medium = Heat lost by water

= ( Mcp ( To - 6 ) = Mcp ( 30 - To )

since the properties of water and the cooling medium ( water ) is the same

= 0.02 ( To - 6 ) = 0.011 ( 30 - To )

= 1.82 ( To - 6 ) = 30 - To

hence To ( outlet temperature ) = 14.506°C

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3 years ago

Kim is working on the cost estimate and feasible design options for a building. Which stage of a construction plan is Kim workin

melisa1 [442]

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4 years ago

Which vegetation type do you think will cause fire to spread the fastest?

Ivan

Answer:B

Explanation: Dead Grass more flammable than wood and Green grass contains water.

7 0

3 years ago

Integrate <br>∫cos²x sinx dx<br>

Marianna [84]

Answer:

-⅓ cos³ x + C

Explanation:

∫ cos² x sin x dx

If u = cos x, then du = -sin dx.

∫ -u² du

Integrate using power rule:

-⅓ u³ + C

Substitute back:

-⅓ cos³ x + C

3 0

4 years ago

A rigid copper tank, initially containing 1 m3 of air at 295 K, 5 bar, is connected by avalve to a large supply line carrying ai

asambeis [7]

Answer:

Initial mass = 5.91 kg

Final mass = 16.8 kg

Heat transfer Q = - 625.9 KJ

Explanation:

Given data

Tank volume V = 1 $m^{3}$

Entering outside air temperature $T_{i} = 295 K$

Entering outside air pressure $P_{i}$ = 15 bar

Initial tank pressure $P_{1}$ = 5 bar

Initial tank temperature $T_{1}$ = 295 K

Final pressure $P_{2}$ = 15 bar

Final temperature $T_{2}$ = 310 K

We know that

P V = m R T

(a). Initial mass is given by

$m_{1} = \frac{P_{1} V_{1} }{R T_{1} }$

Put all the values in given equation

$m_{1} = \frac{(500000)(1)}{(287)(295)} = 5.91 \ kg$

(b). Final mass is given by

$m_{2} = \frac{(1500000)(1)}{(287)(310)}$

$m_{2} = 16.8 \ kg$

This is the final volume of the tank.

Δ U = Δ q + $h_{i}$ Δ $m_{cv}$

$m_{2} u_{2} - m_{1} u_{1} = Q + h_{i} (m_{2} - m_{1} )$

Specific internal energy at initial temperature & pressure $u_{1}$ = 210.5 $\frac{KJ}{Kg}$

Specific internal energy at final temperature & pressure $u_{2}$ = 221.25 $\frac{KJ}{Kg}$

Specific enthalpy is $h_{i} =$ 295.17 $\frac{KJ}{Kg}$

Q = ( 16.8 × 210.48 - 5.91 × 210.49 )- 295.17 ( 16.8 - 5.91 )

Q = 2292.23 - 3214.4

$Q_{a}$ = - 741.4 KJ

The heat transfer for the tank is given by

$Q_{t}$ = m C $(T_{2} - T_{1})$

$Q_{t}$ = 20 × 0.385 × ( 310-295 )

$Q_{t}$ = + 115.5 KJ

Total heat transfer Q = $Q_{a}$ + $Q_{t}$

Q = - 741.4 + 115.5

Q = - 625.9 KJ

This is the heat transfer to the surrounding from the tank.

6 0

3 years ago