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Lerok [7]
3 years ago
11

An immersion heater has a resistance of 50Ω and carries a current of 2.5A current. What will be the final temperature of 500 g o

f water that is initially at 20ºC after 3 minutes if Does the water absorb all the heat given off by the heater? Does the water absorb 40% of the heat released?
Engineering
1 answer:
Lorico [155]3 years ago
8 0

Answer:

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An immersion heater has a resistance of 50Ω and carries a ...brainly.com › Engineering › High School

An immersion heater has a resistance of 50Ω and carries a current of 2.5A current. ... What will be the final temperature of 500 g of water that is initially at 20ºC after 3 minutes if Does the water absorb all the heat given off by the heater? Does ...

Explanation:

Search Results

Web results

An immersion heater has a resistance of 50Ω and carries a ...brainly.com › Engineering › High School

An immersion heater has a resistance of 50Ω and carries a current of 2.5A current. ... What will be the final temperature of 500 g of water that is initially at 20ºC after 3 minutes if Does the water absorb all the heat given off by the heater? Does ...

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a) The required additional minterms  for f so that f has eight primary implicants with two literals and no other prime implicant are m_{2},m_{3},m_{7},m_{8},m_{11},m_{12},m_{13},m_{14} and m_{15}

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A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is igno
olga55 [171]

Answer:

critical clearing angle = 70.3°

Explanation:

Generator operating at = 50 Hz

power delivered = 1 pu

power transferable when there is a fault = 0.5 pu

power transferable before there is a fault = 2.0 pu

power transferable after fault clearance = 1.5 pu

using equal area criterion to determine the critical clearing angle

Attached is the power angle curve diagram and the remaining part of the solution.

The power angle curve is given as

= Pmax sinβ

therefore :  2sinβo = Pm

                   2sinβo = 1

                   sinβo = 0.5 pu

                   βo = sin^{-1} (0.5) = 30⁰

also ;   1.5sinβ1 = 1

               sinβ1 = 1/1.5

               β1 = sin^{-1} (\frac{1}{1.5} ) = 41.81⁰

∴ βmax = 180 - 41.81  = 138.19⁰

attached is the remaining solution

The critical clearing angle = cos^{-1} 0.3372  ≈ 70.3⁰

3 0
3 years ago
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