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3 years ago

Which statements describe a closed circuit? Check all that apply.

Chemistry

2 answers:

Verdich [7]3 years ago

8 0

The answer is:

"Bulbs will shine.

The circuit is complete.

Charges flow."

I just took the test and got it right, good luck!

ehidna [41]3 years ago

7 0

Answer:

The answers are options A, D and E

Explanation:

Shut circuit implies a total electrical association around which flow streams or courses. When you have a progression of electrical wires interfacing with one another and finishing a circuit so flow goes from one end of the hover to alternate, this is a case of a shut circuit. A broken wire or an "open" (off) switch both leave holes in a circuit keeping electrons from going from one side of the power source to the next. In this way, electrons won't stream. This circumstance is called an open circuit. A shut (on) switch implies that the circuit through the switch is connected.The meaning of an open circuit is a broken way for an electrical flow because of an open switch or frayed wire. A case of an open circuit is a string of electric lights that don't work on the off chance that one globule goes out. YourDictionary definition and utilization precedent. Shut circuit implies an entire electrical association around which flow streams or circles. When you have a progression of electrical wires associating with one another and finishing a circuit so flow goes from one end of the hover to alternate, this is a case of a shut circuit.

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Read 2 more answers

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/K

mina [271]

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: