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3 years ago

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Select the correct statement to describe when a sample of liquid water vaporizes into water vapor. Question 12 options: Temperat

ure increases and molecular motion increases while shape becomes less defined. Temperature increases and molecular motion decreases while shape becomes more defined. Temperature decreases and molecular motion decreases while shape becomes more defined. Temperature decreases and molecular motion increases while shape becomes less defined.

1 answer:

pashok25 [27]3 years ago

3 0

<span>temperature increases and molecular motion increases while shape becomes less defined.

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A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw

mezya [45]

Answer:

The mass is $m = 3.45408 kg$

Explanation:

From the question we are told that

The extension of the rod is $from , \ x_1 = 0 \to x_2 = 13.0$

The area is $A = 8.0 cm^2$

The density increase as follows $from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3$

The equation $\rho = B + Cx$

at $x_1= 0$ $\rho_1 =2.5 g/cm^2$

So

$2.5 = B + 0$

=> $B =2.5$

So at $x_2 = 13.0$ , $\rho_2= 19.0 g/cm^3$

So

$19.0 = 2.5 + C(13)$

=> $C = 1.27$

Now

$m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx$

$m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.$

$m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ]$

$m = 3454.08 g$

$m = 3.45408 kg$

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3 years ago

To receive AM radio, you want an RLC circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is acc

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The formula for resonant frequency is:

$f_0=\frac{1}{2\pi\sqrt{LC} }$

Given information:

$f_{0\text{,small}}=500 \text{ kHz}\\f_{0\text{,large}}=1650 \text{ kHz}\\L=3.83\text{ } \mu \text{H}$

Plug in the given values to find one value of capacitance:

$500 \text{ kHz}=\frac{1}{2\pi\sqrt{C(3.83\text{ } \mu \text{H})} }\\C=2.645*10^{-8} \text{ F}=26.45 \text{ nF}$

Plug in the given values to find the other value of capacitance:

$1650 \text{ kHz}=\frac{1}{2\pi\sqrt{C(3.83\text{ } \mu \text{H})} }\\C=2.429*10^{-8} \text{ F}=2.429 \text{ nF}$

This gives a range of 2.429 nF to 26.45 nF.

With significant figures taken into account, the range of capacitance is 2.43 nF to 30 nF.

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