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Simora [160]
3 years ago
7

In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o

f this change on the impedance of the circuit

Physics
1 answer:
likoan [24]3 years ago
3 0

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

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How much time would it take for an object to fall 4.7 meters
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Answer:

4.7 is 10 as much as the number 0.47.

If you multiply 0.47 x 10 it will equal 4.7

Explanation:

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2 years ago
Find the magnitude of the free-fall acceleration at the orbit of the Moon (a distance of 60RE from the center of the Earth with
Ede4ka [16]

Answer:

The magnitude of the free-fall acceleration at the orbit of the Moon is 2.728\times 10^{-3}\,\frac{m}{s^{2}} (\frac{2.784}{10000}\cdot g, where g = 9.8\,\frac{m}{s^{2}}).

Explanation:

According to the Newton's Law of Gravitation, free fall acceleration (g), in meters per square second, is directly proportional to the mass of the Earth (M), in kilograms, and inversely proportional to the distance from the center of the Earth (r), in meters:

g = \frac{G\cdot M}{r^{2}} (1)

Where:

G - Gravitational constant, in cubic meters per kilogram-square second.

M - Mass of the Earth, in kilograms.

r - Distance from the center of the Earth, in meters.

If we know that G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, M = 5.972\times 10^{24}\,kg and r = 382.26\times 10^{6}\,m, then the free-fall acceleration at the orbit of the Moon is:

g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}

g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}

6 0
2 years ago
A 440 kg roller coaster car is going 26 m/s when it reaches the lowest point on the track. If the car started from rest at the t
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6 0
2 years ago
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Describe Earth's three global wind belts
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4 0
2 years ago
A 43-kg child sits in a massless swing. With what horizontal force must the seat be pulled so that the ropes form an angle of 35
GenaCL600 [577]

Answer:

F_x = 295.4N

Explanation:

180°-90°-35°= 55°

F= m(a)

F_y = 43kg(9.81m/s²)

F_y = 421.83N

tan(55°) = 421.83N/F_x

F_x = 421.83N/tan(55°)

F_x = 295.3685458N

7 0
2 years ago
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