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3 years ago

Lahat ay tumutukoy sa bugtong, maliban sa:

Physics

1 answer:

AlladinOne [14]3 years ago

5 0

Answer:

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Explanation:

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A series RLC circuit with a resistance of 121.0 Ω has a resonance angular frequency of 5.1 ✕ 105 rad/s. At resonance, the voltag

Vadim26 [7]

Answer:

Explanation:

At resonance ω₀L = 1 / ω₀C , L is inductance and C is capacitance .

C = 1 / ω₀²L , ω₀ = 5.1 x 10⁵ . ( given )

voltage over resistance = R I , R is resistance and I is current

voltage over inductance = Iω₀L

R I / Iω₀L = 60 / 40

R / ω₀L = 3 / 2

L = 2 R / 3 ω₀

= 2 x 121 / 3 x 5.1 x 10⁵

= 15.81 x 10⁻⁵

C = 1 / ω₀²L

= 1 / (5.1 x 10⁵)² x 15.81 x 10⁻⁵

= .002432 x 10⁻⁵

= 24.32 x 10⁻⁹ F

Let the angular frequency required be ω

Tan 45 = (ωL - 1 / ωC) / R

ωL - 1 / ωC = R

ω²LC - 1 = R ωC

ω²LC = 1 + R ωC

ω² x 15.81 x 10⁻⁵ x 24.32 x 10⁻⁹ = 1 + 121 x ω x 24.32 x 10⁻⁹

ω² x 384.5 x 10⁻¹⁴ = 1 + 2942.72 x10⁻⁹ω

ω² - 7.65 x 10⁶ ω - 1 = 0

ω = 7.65 x 10⁶

frequency = 7.65 x 10⁶ / 2π

= 1.22 x 10⁶ Hz

3 0

3 years ago

How are force, mass and acceleration related?

Zepler [3.9K]

Newton's 2nd law of motion:

F = ma

F is force, m is mass, and a is acceleration.

6 0

3 years ago

A rock is thrown downward from the top of a 36.2-m-tall tower with an initial speed of 13 m/s. Assuming negligible air resistanc

vekshin1

Explanation:

u = 13m/s

s = 36.2m

a = 9.81m/s² (positive because it is acting on the same direction as the velocity)

Using Kinematics, we have v² = u² + 2as = (13)² + 2(9.81)(36.2) = 879.244m²/s². => v = 29.65m/s.

4 0

3 years ago

What is the reading shown in millimetre

Kitty [74]

Answer:

3mm

Explanation:

4 0

2 years ago

State and prove bessel inequality

maria [59]

Statement :- We assume the orthagonal sequence ${{\{\phi\}}_{1}^{\infty}}$ in Hilbert space, now ${\forall \sf \:v\in \mathbb{V}}$ , the Fourier coefficients are given by:

${\quad \qquad \longrightarrow \sf a_{i}=(v,{\phi}_{i})}$

Then Bessel's inequality give us:

${\boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Proof :- We assume the following equation is true

${\quad \qquad \longrightarrow \displaystyle \sf v_{n}=\sum_{i=1}^{n}a_{i}{\phi}_{i}}$

So that, ${\bf v_n}$ is projection of ${\bf v}$ onto the surface by the first ${\bf n}$ of the ${\bf \phi_{i}}$ . For any event, ${\sf (v-v_{n})\perp v_{n}}$

Now, by Pythagoras theorem:

${:\implies \quad \sf \Vert v\Vert^{2}=\Vert v-v_{n}\Vert^{2}+\Vert v_{n}\Vert^{2}}$

${:\implies \quad \displaystyle \sf ||v||^{2}=\Vert v-v_{n}\Vert^{2}+\sum_{i=1}^{n}\vert a_{i}\vert^{2}}$

Now, we can deduce that from the above equation that;

${:\implies \quad \displaystyle \sf \sum_{i=1}^{n}\vert a_{i} \vert^{2}\leqslant \Vert v\Vert^{2}}$

For ${\sf n\to \infty}$ , we have

${:\implies \quad \boxed{\displaystyle \bf \sum_{1}^{\infty}\vert a_{i}\vert^{2}\leqslant \Vert v\Vert^{2}}}$

Hence, Proved

5 0

2 years ago

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