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klio [65]
4 years ago
10

Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is abo

ut f = 4.11 x 1012 Hz, and the amplitude is about 1.23 x 10^-11 m. For a typical atom, what is its (a) maximum speed and (b) maximum acceleration
Physics
1 answer:
hoa [83]4 years ago
4 0

To develop this problem it is necessary to use the equations of description of the simple harmonic movement in which the acceleration and angular velocity are expressed as a function of the Amplitude.

Our values are given as

f = 4.11 *10^{12} Hz

A = 1.23 * 10^{-11}m

The angular velocity of a body can be described as a function of frequency as

\omega = 2\pi f

\omega = 2\pi 4.11 *10^{12}

\omega=2.582*10^{13} rad/s

PART A) The expression for the maximum angular velocity is given by the amplitude so that

V = A\omega

V =( 1.23 * 10^{-11})(2.582*10^{13})

V =  = 317.586m/s

PART B) The maximum acceleration on your part would be given by the expression

a = A \omega^2

a =( 1.23 * 10^{-11})(2.582*10^{13})^2

a= 8.2*10^{15}m/s^2

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A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of ki
Neporo4naja [7]

Answer:

a = -1.961\,\frac{m}{s^{2}}, s = 16.318\,m, t = 4.079\,s

Explanation:

The equations of equilibrium for the rock are:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a

\Sigma F_{y} = N - m\cdot g = 0

After some algebraic handling, the following expression is found:

-\mu_{k}\cdot m \cdot g = m \cdot a

-\mu_{k}\cdot g = a

Deceleration experimented by the rock is:

a = - (0.2)\cdot (9.807\,\frac{m}{s^{2}} )

a = -1.961\,\frac{m}{s^{2}}

The distance travelled by the rock before stopping is:

s = \frac{(0\,\frac{m}{s} )^{2}-(8\,\frac{m}{s} )^{2}}{2\cdot (-1.961\,\frac{m}{s^{2}} )}

s = 16.318\,m

And the time is:

t = \frac{0\,\frac{m}{s}-8\,\frac{m}{s}}{(-1.961\,\frac{m}{s^{2}} )}

t = 4.079\,s

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I think it will be B
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If a power utility were able to replace an existing 500 kV transmission line with one operating at 1 MV, it would change the amo
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Answer:

It would change the amount of heat produced in the transmission line to four times the previous value.

Explanation:

Given;

initial voltage in the transmission line, V₁ = 500 kV = 500,000 V

Final voltage in the transmission line, V₂ = 1 MV = 1,000,000

The power lost in the transmission line due to heat is given by;

P = \frac{V^2}{R}

Power lost in the first wire;

P_1 = \frac{V_1^2}{R}

R = \frac{V_1^2}{P_1}

Power lost in the second wire

P_2 = \frac{V_2^2}{R}\\\\ R = \frac{V_2^2}{P_2}

Keeping the resistance constant, we will have the following equation;

\frac{V_2^2}{P_2} = \frac{V_1^2}{P_1} \\\\P_2 = \frac{V_2^2P_1}{V_1^2}\\\\

P_2 = \frac{(1,000,000)^2P_1}{(500,000)^2}\\\\P_2 =4P_1

Therefore, it would change the amount of heat produced in the transmission line to four times the previous value.

8 0
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