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In-s [12.5K]
3 years ago
7

A gas at constant temperature is confined to a cylinder with a movable piston. The piston is slowly pushed into the cylinder, de

creasing the volume of a gas. The pressure increases. What are the variables in this experiment? (What is changing?)
a. temperature and time
b. time and volume
c. volume and pressure
d. pressure and temperature
Physics
2 answers:
Inessa [10]3 years ago
4 0
C. Volume and pressure
Key words: increases and decreases
Romashka [77]3 years ago
4 0
The answer is C. Volume and pressure
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A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci
Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = \frac{C_{3}C_{4}  }{C_{3} + C_{4}  }

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = \frac{4.25\times3.98 }{4.25 + 3.98  }

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3  = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = \frac{Q}{C_{4} } = \frac{35.46\times10^{-6} }{3.98\times10^{-6}} = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0
3 years ago
Receiver maxima problem. When the receiver moves through one cycle, how many maxima of the standing wave pattern does the receiv
shusha [124]

Answer:

Two.

Explanation:

  • A standing wave or stationary wave are the type of wave in which it oscillates with time but the amplitude does not move.
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A loop of wire is placed in a uniform magnetic field. (a) For what orientation of the loop is the magnetic flux a maximum? The p
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Answer:

1) The plane of the loop is perpendicular to the magnetic field.

2) The magnetic flux is independent of the orientation of the loop.p

Explanation:

The flux is calculated as φ=BAcosθ. The flux is therefore the highest when the magnetic field vector is perpendicular to the plane of the loop We can also deduce that the flux is zero when there is no magnetic field part perpendicular to the loop When the angle reaches zero, the flux is in the limit because when the angle becomes zero, the cos is the maximum.

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To calculate the velocity relative to car A, you have to add the opposite of the velocity of car A to car B.

the velocity of car B relative to car A:
(let east and north directions be positive)
v = v (car B) - v(car A)
v(east-west direction) = -120 * cos 30° - 110
v(north - south direction) = -120 * sin 30° - 0
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