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3 years ago

13

If a string vibrates at the fundamental frequency of 528 Hz and also produces an overtime with a frequency of 1,056 Hz this over

tune is the

1 answer:

Romashka-Z-Leto [24]3 years ago

5 0

I have a hunch that you're not talking about overtime
OR overtune. I think you're going for overtone .

For a fundamental frequency of 528 Hz,
1,056 Hz is the second harmonic.

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A yoyo with a mass of m = 150 g is released from rest as shown in the figure.

avanturin [10]

(1) The linear acceleration of the yoyo is 3.21 m/s².

(2) The angular acceleration of the yoyo is 80.25 rad/s²

(3) The weight of the yoyo is 1.47 N

(4) The tension in the rope is 1.47 N.

(5) The angular speed of the yoyo is 71.385 rad/s.

<h3> Linear acceleration of the yoyo</h3>

The linear acceleration of the yoyo is calculated by applying the principle of conservation of angular momentum.

∑τ = Iα

rT - Rf = Iα

where;

I is moment of inertia
α is angular acceleration
T is tension in the rope
r is inner radius
R is outer radius
f is frictional force

rT - Rf = Iα ----- (1)

T - f = Ma -------- (2)

a = Rα

where;

a is the linear acceleration of the yoyo

Torque equation for frictional force;

$f = (\frac{r}{R} T) - (\frac{I}{R^2} )a$

solve (1) and (2)

$a = \frac{TR(R - r)}{I + MR^2}$

since the yoyo is pulled in vertical direction, T = mg $a = \frac{mgR(R - r)}{I + MR^2} \\\\a = \frac{(0.15\times 9.8 \times 0.04)(0.04 - 0.0214)}{1.01 \times 10^{-4} \ + \ (0.15 \times 0.04^2)} \\\\a = 3.21 \ m/s^2$

<h3>Angular acceleration of the yoyo</h3>

α = a/R

α = 3.21/0.04

α = 80.25 rad/s²

<h3>Weight of the yoyo</h3>

W = mg

W = 0.15 x 9.8 = 1.47 N

<h3>Tension in the rope </h3>

T = mg = 1.47 N

<h3>Angular speed of the yoyo </h3>

v² = u² + 2as

v² = 0 + 2(3.21)(1.27)

v² = 8.1534

v = √8.1534

v = 2.855 m/s

ω = v/R

ω = 2.855/0.04

ω = 71.385 rad/s

Learn more about angular speed here: brainly.com/question/6860269

#SPJ1

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1 year ago

Calculate the number of vacancies per cubic meter in gold (au) at 900c. the energy for vacancy formation is 0.98 ev/atom. furthe

Sergeeva-Olga [200]

My mom said its f so i think its q

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2 years ago

Which of the following statements describes the trend in the data?

AURORKA [14]

Answer:

Explanation:

The answer is D- The number of flies decreased, then leveled off.

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2 years ago

What would the final volume of 40 L of gas at 80 pascals be if the pressure increases to 130 pascals?

melomori [17]

Answer:

Final volume, V2 = 24.62 L

Explanation:

Given the following data;

Initial volume = 40 L

Initial pressure = 80 Pa

Final pressure = 130 Pa

To find the final volume V2, we would use Boyles' law.

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by;

$PV = K$

$P_{1}V_{1} = P_{2}V_{2}$

Substituting into the equation, we have;

$80 * 40 = 130V_{2}$

$3200 = 130V_{2}$

$V_{2} = \frac {3200}{130}$

$V_{2} = 24.62$

Final volume, V2 = 24.62 Liters

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2 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago