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2 years ago

7

Suppose a diving board with no one on it bounces up and down in a SHM with a frequency of 4.00 Hz. The board has an effective ma

ss of 10.0 kg. What is the frequency of the SHM of a 75.0-kg diver on the board?

1 answer:

kirza4 [7]2 years ago

7 0

Answer:

I took 3*sqrt(10/83)= 1.110349815

And rounded to 1.11 Hz

Explanation:

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The distance is 28 meters, and the displacement is -4.

For the distance it would be 12 + 16 = 28.

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2 years ago

Which group contains elements that have the following characteristics:

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2 years ago

A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an

olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = $\pi r^2$

V = Potential applied = 2 mV

k = Dielectric constant = 40

$\epsilon_0$ = Electric constant = $8.854\times 10^{-12}\ \text{F/m}$

Capacitance is given by

$C=\dfrac{k\epsilon_0A}{d}$

Charge is given by

$Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}$

Number of electron is given by

$n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}$

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

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2 years ago

In an internal combustion engine, heat flow into a gas causes it to <br> .

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Answer:

Along path BC of the Otto cycle, heat transfer Qh into the gas occurs at constant volume, causing a further increase in pressure and temperature. This process corresponds to burning fuel in an internal combustion engine, and takes place so rapidly that the volume is nearly constant.

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1 year ago

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r

andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature $T_1=781 K$

Reservoir 2 Temperature $T_2=335 K$

Let Q is the amount of heat Flows i.e. $Q=1477 J$

thus change in Entropy is given by $\frac{\sum Q}{T}$

$\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}$

$\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}$

$\Delta S=\frac{\sum Q}{T}=-1.891+4.4089$

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2 years ago