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3 years ago

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(a) When the displacement in SHM is one-sixth the amplitude xm, what fraction of the total energy is kinetic energy? (b) What fr

action is potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy? (Use any variable or symbol stated above as necessary.)

1 answer:

andreev551 [17]3 years ago

3 0

Answer:

Explanation:

$KE = \frac{1}{2}\times K\times (A^2-X^2)$

at $X = \frac{1}{6}A\\\\
KE = \frac{1}{2}\times K\times ( A^2 - A^{\frac{2}{36})\\\\
fraction = \frac{35}{36} = 0.9722
$

b) PE fraction = 1/36 = 0.02777

c) 1/2*KX^2 = 1/2*K*(A^2-X^2)

=> X = A/sqrt(2)

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If the planes of a crystal are 3.50 (1 A= 10^-10m = Ångstrom unit) apart, what wavelength of electromagnetic waves are needed so

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Answer:

λ = 2.62 x 10⁻¹⁰ m = 0.262 nm

Explanation:

We can use Bragg's Law's equation to solve this problem. The Bragg's Law's equation is written as follows:

mλ = 2d Sin θ

where,

m = order of reflection = 1

λ = wavelength = ?

d = distance between the planes of crystal = 3.5 x 10⁻¹⁰ m

θ = strike angle of waves on plane = 22°

Therefore, substituting the respective values in the equation, we get:

(1)λ = (2)(3.5 x 10⁻¹⁰ m)(Sin 22°)

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A force of 1000 newtons was necessary to lift a rock. A total of 3000 joules of work was done. How far was the rock lifted?

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Answer:

Explanation:

3 meters

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2 years ago

The car salesman tells you that the car can go from a stop position to 60 mph in six seconds is giving you the car’s rate of

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The salesman is telling you the average magnitude of the car's acceleration.

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4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3

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Expand each vector into their component forms:

$\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N$

Similarly,

$\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N$

$\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N$

Then assuming the resultant vector $\vec R$ is the sum of these three vectors, we have

$\vec R=\vec A+\vec B+\vec C$

$\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N$

and so $\vec R$ has magnitude

$\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N$

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3 years ago

In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for

4vir4ik [10]

Answer:

The fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is

$k = 0.903$

Explanation:

From the question we are told that

The time constant $\tau = 3$

The potential across the capacitor can be mathematically represented as

$V = V_o (1 - e^{- \tau})$

Where $V_o$ is the voltage of the capacitor when it is fully charged

So at $\tau = 3$

$V = V_o (1 - e^{- 3})$

$V = 0.950213 V_o$

Generally energy stored in a capacitor is mathematically represented as

$E = \frac{1}{2 } * C * V ^2$

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now since capacitance is constant at $\tau = 3$

The energy stored can be evaluated at as

$V^2 = (0.950213 V_o )^2$

$V^2 = 0.903 V_o ^2$

Hence the fraction of the energy stored in an initially uncharged capacitor is

$k = 0.903$

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3 years ago