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Lana71 [14]
1 year ago
6

A 2-f and a 1-f capacitor are connected in series and a voltage is applied across the combination. the 2-f capacitor has:_______

.
Physics
2 answers:
VARVARA [1.3K]1 year ago
7 0

Half the potential difference of the the1-µF

A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.

The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.

Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.

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LenKa [72]1 year ago
3 0

Half the potential difference of the 1μf capacitor.

Charge: Q = CV where C is the capacitance in Farads, V is the voltage across the capacitor in Volts and Q is the charge measured in coulombs (C).

Energy stored: W = ½ QV = ½ CV×V

where W is the energy measured in Joules.

As the capacitance of a capacitor is equal to the ratio of the stored charge to the potential difference across its plates, giving: C = Q/V, thus V = Q/C as Q is constant across all series connected capacitors, therefore the individual voltage drops across each capacitor is determined by its its capacitance value.

When the capacitors are connected in the form of series combination, the capacitance in total is less than the individual capacitances of the series capacitors. If one, two or number of capacitors are connected in the series form, the overall effect is the single or equivalent capacitor which has the total sum of the spacings between the plates of the individual capacitors. The increase in the plate spacing results in the decreased capacitance, with all the other factors remaining unchanged.

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A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 0
3 years ago
You indicate that a symbol
Goryan [66]

Answer:

B. over the symbol.

Explanation:

vectors are represented with a symbol carrying an arrow head with also indicates direction

8 0
3 years ago
Bryce, a mouse lover, keeps his four pet mice in a roomy cage, where they spend much of their spare time–when they\'re not sleep
Anit [1.1K]

Answer:

mice total momentum (-0.000250, 0.00639) Kg m

Explanation:

To calculate the moment of the mice we must multiply their mass by their velocities, remember that the moment is a vector quantity, so we use the components of velocity

mouse 1

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  V1 = (0.869, -0.283) m / s

 

  Px = m Vx

  Px1 = 0.0225 0.869

  Px1 = 0.01955 Kg m

  Py = m Vy

  Py1 = 0.0225 (-0.283)

  Py1 = -0.006368 Kg m

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Mouse 2

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 P2 = (-0.0196, -0.00564) Kg m

Mouse 3

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Mouse4

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To find the total momentum we must add each component of the individual moments

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When you changed from low to high power, how did the change affect the working distance of the lens?
Basile [38]

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Magnification change The magnification of a specimen is increased by switching from low power to high power. The magnification of an image is determined by multiplying the magnification of the objective lens by the magnification of the ocular lens, or eyepiece.

The geometry of the optical system connects the magnifying power, or how much the thing being observed seems expanded, and the field of view, or the size of the object that can be seen.

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brainly.com/question/13551539

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Less than the distance

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