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jek_recluse [69]

3 years ago

13

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

4.42 Hz. The amplitude of the motion is 8.26 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring.
(a) What is the amplitude and
(b) the frequency of the simple harmonic motion that exists after the block splits?

1 answer:

Nastasia [14]3 years ago

8 0

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

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Part A)
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Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
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Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

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$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

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Answer:

105.8 m

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Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

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To solve the problem it is necessary to apply the concepts related to Byle's Law and Avogadro's Law.

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R = Universal as constant

n = number of moles

V = Volume

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For our case we have that the mass of Zn is 2.2g in moles would be

$[tex]Zn = \frac{2.2}{65}$ [/tex]

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We know that 1 mole of hydrogen gas is proceed by 1 mole of zinc and the result is $Zn^{2+}$ , then Hydrogen can produce the same quantity,

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Applying the previous equation we have that

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exis [7]

Answer:

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Explanation:

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7 0

3 years ago