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Vsevolod [243]
3 years ago
14

This circuit is?

Engineering
2 answers:
torisob [31]3 years ago
5 0
Correct answer is open and on
frez [133]3 years ago
4 0

Answer:

The circuit is open and on.

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What is an aqueduct?
Lady bird [3.3K]
Hi there!

1. C) Ancient Structure

2. A) Iterate

3. A) NYC

The first two, I’m sure the third i’m not.

Sorry if I’m wrong and hope this helps !
6 0
4 years ago
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Suppose you must remove an average of 3.9×108J of thermal energy per day to keep your house cool during the summer. Part A If yo
noname [10]

Answer:

The required mechanical work is required to reduce each day by 1.05×10^8 Joules.

Explanation:

Coefficient of Performance (COP) = Q/W

Q is thermal energy absorbed by the air conditioner

W is mechanical work done

Q = 3.9×10^8 J

COP of old air conditioner = 2.3

W = Q/COP = 3.9×10^8/2.3 = 1.70×10^8 J

COP of new air conditioner = 6

W = Q/COP = 3.9×10^8/6 = 6.5×10^7 J

Reduction in mechanical work = (1.7×10^8) - (6.5×10^7) = 1.05×10^8 J

3 0
3 years ago
A ladder logic program and the associated physical input/output components are given below. Lighting changes from full darkness
Katena32 [7]

Answer:

O2 is true.

Explanation:

8 0
4 years ago
This operating mode assists steering return after completing a turn.
castortr0y [4]

Answer:

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3 0
2 years ago
A rigid canister with a radius of 5 in and a height of 10 in is filled with air. The initial pressure and temperature of air in
Elza [17]

Answer:1.458 Btu

Explanation:

Given

radius of canister\left ( r\right )=5in

Height of canister\left ( h\right )=10 in.

Initial pressure\left ( P_i\right )=14.7 Psi

Initia ltemprature\left ( T_i\right )=70^{\circ}F

Final pressure\left ( P_f\right )=30Psi

as canister is rigid therefore change in volume is zero

therefore

\frac{P_i}{T_i}=\frac{P_f}{T_f}

\frac{14.7}{70}=\frac{30}{T_f}

T_f=142.85^{\circ}F

volume of canister=\pi \times r^{2}\times h

                               =\pi \times 5^{2}\times 10=250\pi

volume of canister=12872,038.8 mm^3

now calculating mass of air

PV=mRT

substituting values

\left ( 14.7Psi\right )\left ( 12872,038.8 mm^3\right )=m\left ( 0.287\right )\left ( 70^{\circ}F\right )

m=21.0192 gm

Therefore heat transferred =mc_p\left ( T_f-T_i\right )

                                             =21.0192\times 10^{-3}\times \left ( 142.857-70\right )

                                             =1539.052J=1.458Btu

3 0
4 years ago
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