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MrRissso [65]
4 years ago
11

How could you increase power of a wave in a spring

Physics
2 answers:
lora16 [44]4 years ago
8 0
Alll of the above I just read it from someone and if it’s not that lmk
coldgirl [10]4 years ago
6 0

any of those or all

Increase the amplitude

Increase the frequency

Increase the tension on the spring

Explanation:

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What problem do refractor telescopes have that reflectors don't? Group of answer choices bad seeing light loss from secondary el
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chromatic aberration problem do refractor telescopes have that reflectors don't

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Chromatic aberration is a phenom in which light rays crossing through a lens focus at various points, depending on their wavelength. Chromatic aberration is a dilemma in which lens or refracting, telescopes undergo from. The various image distances for the respective colors affect various image sizes for them.

This involves the creation of disturbing color fringes in the image. Chromatic aberration can be pretty well adjusted by the use of an achromatic doublet. Here, a positive biconvex lens is coupled with a negative lens placed backward with greater dispersion. Thus partly compensates for the chromatic aberration.

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The force of attraction between 2 objects due to their masses is called what?
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Answer:

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Which of these activities increases the amount of carbon in the atmosphere?
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3 years ago
A 10.0-µF capacitor is charged so that the potential difference between its plates is 10.0 V. A 5.0-µF capacitor is similarly ch
IceJOKER [234]

Answer:

Explanation:

Given that,

First Capacitor is 10 µF

C_1 = 10 µF

Potential difference is

V_1 = 10 V.

The charge on the plate is

q_1 = C_1 × V_1 = 10 × 10^-6 × 10 = 100µC

q_1 = 100 µC

A second capacitor is 5 µF

C_2 = 5 µF

Potential difference is

V_2 = 5V.

Then, the charge on the capacitor 2 is.

q_2 = C_2 × V_2

q_2 = 5µF × 5 = 25 µC

Then, the average capacitance is

q = (q_1 + q_2) / 2

q = (25 + 100) / 2

q = 62.5µC

B. The two capacitor are connected together, then the equivalent capacitance is

Ceq = C_1 + C_2.

Ceq = 10 µF + 5 µF.

Ceq = 15 µF.

The average voltage is

V = (V_1 + V_2) / 2

V = (10 + 5)/2

V = 15 / 2 = 7.5V

Energy dissipated is

U = ½Ceq•V²

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U = 4.22 × 10^-4 J

U = 422 × 10^-6

U = 422 µJ

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4 years ago
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