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3 years ago

The note "Middle C" is known to have a frequency of 261.6 Hz. What would

Physics

1 answer:

Gwar [14]3 years ago

8 0

Answer: frequency = 523.2 Hz

Explanation:

One octave higher = 2f of the last octave.

For instance, if C2 = 4 Hz

A octave higher will be C3 in which its frequency will be 2 × 4Hz = 8Hz

Back to the question.

One octave higher frequency will therefore equal to:

2f = 2 × 261.6 = 523.2 Hz

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FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

Please someone help me

beks73 [17]

New substances are formed by chemical reactions. When elements react together to form compounds their atoms join to other atoms using chemical bonds. For example, iron and sulfur react together to form a compound called iron sulfide. Hopefully this will help you decide...

5 0

3 years ago

What is the source of all magnetism?

ivanzaharov [21]

The answer is electric current

3 0

3 years ago

A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Natasha2012 [34]

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0

3 years ago

The block shown above has a mass of 105 g.<br> What is the density of the block? SC.8.P.8.3

zaharov [31]

The density of the block is 1.25 cm³

The correct answer to the question is Option B. 1.25 cm³

To solve this question, we'll begin by calculating the volume of the block. This can be obtained as follow:

Length = 7 cm

Height = 4 cm

Width = 3 cm

<h3>Volume =? </h3>

Volume = Length × Width × Height

Volume = 7 × 3 × 4

<h3>Volume = 84 cm³</h3>

Thus, the volume of the block is 84 cm³

Finally, we shall determine the density of the block. This can be obtained as follow:

Density is defined as mass per unit volume i.e

$Density = \frac{mass}{volume} \\\\$

Mass of block = 105 g

Volume of block = 84 cm³

<h3>Density of block =? </h3>

$Density = \frac{mass}{volume}\\\\Density = \frac{105}{84}\\\\$

<h3>Density of block = 1.25 cm³</h3>

Therefore, the density of the block is 1.25 cm³.

Hence, Option B. 1.25 cm³ gives the correct answer to the question.

Learn more: brainly.com/question/2040396?referrer=searchResults

4 0

2 years ago