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tensa zangetsu [6.8K]
3 years ago
12

PLEASE HELP !! A 10 ohm resistor is connected to a 9V battery how much current is flowing through the circuit?

Physics
1 answer:
kompoz [17]3 years ago
8 0

Resitance (R)= 10 Ohm

Potential difference (V) = 9V

V= IR

I= V/R

I= 9/10

I= 0.9 Ampere

Therefore 0.9 Ampere of current is flowing through the circuit.

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a * 8 s = 9 m/s
a = 9 m/s : 8 s
a = 1.125 m/s²
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The pathologic changes that occur in the development of coronary atherosclerotic lesions include call damage resulting from whic
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3 effect of oxidized lipids

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Identify three situations in which convection occurs.
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Read 2 more answers
What is the total entropy change if a 0.280 efficiency engine takes 3.78E3 J of heat from a 3.50E2 degC reservoir and exhausts i
babymother [125]

Answer:

\Delta S=1.69J/K

Explanation:

We know,

\eta=1-\frac{T_2}{T_1}=1-\frac{Q_2}{Q_1}      ..............(1)

where,

η = Efficiency of the engine

T₁ = Initial Temperature

T₂ = Final Temperature

Q₁ = Heat available initially

Q₂ = Heat after reaching the temperature T₂

Given:

η =0.280

T₁ = 3.50×10² °C = 350°C = 350+273 = 623K

Q₁ = 3.78 × 10³ J

Substituting the values in the equation (1) we get

0.28=1-\frac{Q_2}{3.78\times 10^{3}}

or

\frac{Q_2}{3.78\times 10^3}=0.72

or

Q_2=3.78\times 10^3\times0.72

⇒ Q_2 =2.721\times 10^3 J

Now,

The entropy change (\Delta S) is given as:

\Delta S=\frac{\Delta Q}{T_1}

or

\Delta S=\frac{Q_1-Q_2}{T_1}

substituting the values in the above equation we get

\Delta S=\frac{3.78\times 10^{3}-2.721\times 10^3 J}{623K}

\Delta S=1.69J/K

7 0
3 years ago
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