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Stolb23 [73]
3 years ago
5

What are 3 common sources of voltage difference?

Physics
1 answer:
rodikova [14]3 years ago
3 0
We are asked to enumerate 3 different sources of potential difference and the answers are listed below:
1. There was two parallel sources and voltage of each source is different from each other.
2. Different batteries are connected.
3, Different values of resistors were used in the circuit.

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A gas has an initial volume of 2.5 L at a temperature of 275 K and a pressure of 2.1 atm. The pressure of the gas increases to 2
olchik [2.2K]

Answer:

2.10L

Explanation:

Given data

V1= 2.5L

T1= 275K

P1= 2.1atm

P2= 2.7 atm

T2= 298K

V2= ???

Let us apply the gas equation

P1V1/T1= P2V2/T2

substitute into the expression we have

2.1*2.5/275= 2.7*V2/298

5.25/275= 2.7*V2/298

Cross multiply

275*2.7V2= 298*5.25

742.5V2= 1564.5

V2= 1564.5/742.5

V2= 2.10L

Hence the final volume is 2.10L

7 0
3 years ago
A balance uses small metal weights of known mass to determine the mass of a substance. Where would a balance NOT function correc
tatuchka [14]

Answer:

D is the answer.

Explanation:

Just do it.

6 0
3 years ago
Read 2 more answers
Why is it necessary to centrifuge out any precipitate formed in the unknown solution and continue testing the remaining unknown
jonny [76]

Answer:

Precipitation is the formation of a solid from a solution. It is necessary to centrifuge the precipitate to exert sufficient forces of gravity to bring the solid particles in the solution to come together and settle

Explanation:

When you centrifuge precipitate it enables the nucleation to form.

Centrifuging the precipitate helps in determining whether a certain element is present in a solution or not.

5 0
3 years ago
What vectors make up R
Tju [1.3M]

Strange as it may seem, I don't think ANY of those choices will work.  Anybody else ?

8 0
3 years ago
A steady device in each cycle converts 155 J of work to thermal energy that is released into a reservoir at 340 K. Part A Calcul
sukhopar [10]

Answer:

change in entropy is 3.3034 × 10^{22}  

Explanation:

give data

thermal energy Q  = 155 J

temperature T = 340 K

to find out

change in entropy

solution

we know change in entropy formula that is

change in entropy = Q / ( K×T )   ..............1

here K is boltzmann constant that is 1.38 × 10^{-23} kg-m²/s²

put these value in equation 1 we get

change in entropy = Q / ( K × T )

change in entropy = 155 / (  1.38 × 10^{-23}  × 340 )

change in entropy = 3.3034 × 10^{22}  

so change in entropy is 3.3034 × 10^{22}  

7 0
3 years ago
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