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kirill115 [55]
3 years ago
5

The amplitude of a lightly damped harmonic oscillator decreases from 60.0 cm to 40.0 cm in 10.0 s. What will be the amplitude of

the harmonic oscillator after another 10.0 s passes?

Physics
1 answer:
Svetllana [295]3 years ago
3 0
This looks like exponential decay starting at 60cm when t is set to zero. See enclosed ...

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valina [46]
Hey there! A pure substance can be a element.

Hope this helps! :)
4 0
4 years ago
Two harmonic waves traveling in opposite directions interfere to produce a standing wave described by y
AnnyKZ [126]

The wavelength of the interfering waves is 3.14 m.

<h3>Calculation:</h3>

The general equation of a standing wave is given by:

y = 2A sin (kx) cos (ωt)                            ......(1)

The given equation represents the standing wave produced by the interference of two harmonic waves:

y = 3 sin (2x) cos 5t                                .......(2)

Comparing equations (1) and (2):

k = 2

We know that,

k = 2π/λ

λ = 2π/k

λ = 2 (3.14)/ 2

λ = 3.14 m

Therefore, the wavelength of the interfering waves is 3.14 m.

I understand the question you are looking for is this:

Two harmonic waves traveling in opposite directions interfere to produce a standing wave described by y = 3 sin (2x) cos 5t where x is in m and t is in s. What is the wavelength of the interfering waves?

Learn more about interfering waves here:

brainly.com/question/2910205

#SPJ4

5 0
2 years ago
A hot-water radiator has a surface temperatue of 80 o C and a surface area of 2 m2 . Treating it as a blackbody, find the net ra
faust18 [17]

Answer:

925.04 J/s

Explanation:

T = 80 C = 80 + 273 = 353 K

To = 20 c = 20 + 273 = 293 K

A = 2 m^2

Use the formula for Stefan's law

Energy radiated per second

E = \sigma  A \left ( T^{4}-T_{0}^{4} \right )

E = 5.67 \times 10^{-8}\times 2\left ( 353^{4}-293^{4} \right )

E = 925.04 J/s

3 0
3 years ago
You take a couple of capacitors and connect them in series, to which you observe a total capacitance
Zepler [3.9K]

Answer:

Approximately \rm 5.7\; \mu F and approximately 29\; \rm \mu F.

Explanation:

Let C_1 and C_2 denote the capacitance of these two capacitors.

When these two capacitors are connected in parallel, the combined capacitance will be the sum of C_1 and C_2. (Think about how connecting these two capacitors in parallel is like adding to the total area of the capacitor plates. That would allow a greater amount of charge to be stored.)

C(\text{parallel}) = C_1 + C_2.

On the other hand, when these two capacitors are connected in series, the combined capacitance should satisfy:

\displaystyle \frac{1}{C(\text{series})} = \frac{1}{C_1} + \frac{1}{C_2}.

(Consider how connecting these two capacitors in series is similar to increasing the distance between the capacitor plates. The strength of the electric field (V) between these plates will become smaller. That translates to a smaller capacitance if the amount of charge stored (Q) stays the same.)

The question states that:

  • C(\text{parallel}) = 35\; \rm \mu F, and
  • C(\text{series}) = 4.8\; \rm \mu F.

Let the capacitance of these two capacitors be x\; \rm \mu F and y\; \rm \mu F. The two equations will become:

\displaystyle \left\lbrace \begin{aligned}& x + y = 35 \\ & \frac{1}{x} + \frac{1}{y} = \frac{1}{4.8}\end{aligned}\right..

From the first equation:

y = 35 - x.

Hence, the y in the second equation here can be replaced with (35 - x). That equation would then become:

\displaystyle \frac{1}{x} + \frac{1}{35 - x} = \frac{1}{4.8}.

Solve for x:

\displaystyle \frac{x + (35 - x)}{x \, (35 - x)} = \frac{1}{4.8}.

x\, (35 - x) = 4.8.

x^2 - 35 \, x + 168 = 0.

Solve this quadratic equation for x:

x \approx 5.7 or x \approx 29.3.

Substitute back into the equation y = 35 - x for y:

  • x \approx 5.7 and y \approx 29.3, or
  • x \approx 29.3 and x \approx 5.7.

In other words, these two capacitors have only one possible set of capacitances (even though the previous quadratic equation gave two distinct real roots.) The capacitances of the two capacitors would be approximately 5.7\; \rm \mu F and approximately 29\; \rm \mu F (both values are rounded to two significant digits.)

6 0
3 years ago
PLEASE HELPPP MEEE :((​
Hoochie [10]
Power = work / time --> time = work / power = 3600 J / 275 watts = 13.1 seconds.

Make me brainliest plz
4 0
3 years ago
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