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sveticcg [70]
3 years ago
15

Determine weather or not the equation below is balanced.

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

Answer:

Yep.. It's balanced and its a combination reaction

Explanation:

Reactants : S₈ + 24F₂

Product 8SF₆

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The male gamete or sperm, is a smaller, mobile cell that meets up with the much larger and less mobile female gamete, egg or ova.

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3 0
3 years ago
The _____ of a wave is the number of wave cycles per second. this quantity is given the symbol ν and has units of s–1 or _____.
Fynjy0 [20]

The frequency of a wave is the number of wave cycles per second. this quantity is given the symbol ν and has units of s^–1 or hertz.

What is a wave ?

A wave is a dynamic disturbance of one or more quantities that propagates through time. When waves oscillate frequently around an equilibrium value at a certain frequency, they are said to be periodic.

What is a wave cycle ?

The wavelength is the measured distance in meters between two waves' peaks or troughs, which is referred to as a wave cycle. It's crucial to remember that the distance from peak to peak equals the distance from trough to trough. The length of time it takes a wave to complete one cycle is known as the wave's period.

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8 0
1 year ago
A parallel-plate air capacitor of area A = 15.0 cm2 and plate separation d = 3.00 mm is charged by a battery to a voltage 58.0 V
sladkih [1.3K]

Answer:

The additional charge that will flow from the battery onto the positive plate is 0.924 nC

Explanation:

Given;

Area of the capacitor, A = 15.0 cm² = 15 x 10⁻⁴ m²

Separation distance, d = 3.00 mm = 3 x 10⁻³ m²

voltage of the capacitor, V = 58.0 V

dielectric constant, k = 4.60

Initial Capacitance of the capacitor before the addition of dielectric material:

C = \frac{\epsilon _oA}{d} = \frac{8.85*10^{-12}*15*10^{-4}}{3*10^{-3}} = 4.425 *10^{-12} F

Initial charge across the parallel plates:

Q₁ = CV

Q = 4.425 x 10⁻¹² x 58 = 2.567 x 10⁻¹⁰ C

Capacitance of the capacitor after the addition of dielectric material:

Cequ. = Ck

Cequ. = 4.425 x 10⁻¹²F x 4.6 = 20.355 x 10⁻¹² F

Final charge across the parallel plates:

Q₂ = Cequ. x V

Q₂ =  20.355 x 10⁻¹² x 58 = 11.806 x 10⁻¹⁰ C

Additional charge = Q₂ - Q₁

                              = 11.806 x 10⁻¹⁰ C - 2.567 x 10⁻¹⁰ C

                              = 9.239 x 10⁻¹⁰ C

                              = 0.924 nC

Therefore, the additional charge that will flow from the battery onto the positive plate is 0.924 nC

3 0
2 years ago
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