Help asap pls. Which of the following best describes how high and low pitch are different?

A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.

Anestetic [448]

Answer:

$\omega=0.37 [rad/s]$

Explanation:

We can use the conservation of the angular momentum.

$L=mvR$

$I\omega=mvR$

Now the Inertia is I(professor_stool) plus mR², that is the momentum inertia of a hoop about central axis.

So we will have:

$(I_{proffesor - stool}+mR^{2})\omega=mvR$

Now, we just need to solve it for ω.

$\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}$

$\omega=\frac{1.5*2.7*0.4}{4.1+1.5*0.4^{2}}$

$\omega=0.37 [rad/s]$

I hope it helps you!

5 0

3 years ago

A particle of mass m moves under an attractive central force F(r) = -Kr4 with angular momentum L. For what energy will the motio

docker41 [41]

Answer:

Angular velocity is same as frequency of oscillation in this case.

ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

Explanation:

- write the equation F(r) = -K $r^{4}$ with angular momentum L

- Get the necessary centripetal acceleration with radius r₀ and make r₀ the subject.

- Write the energy of the orbit in relative to r = 0, and solve for "E".

- Find the second derivative of effective potential to calculate the frequency of small radial oscillations. This is the effective spring constant.

- Solve for effective potential

- ω = $\sqrt{\frac{7K}{m} }$ x $[\frac{L^{2}}{mK}]^{3/14}$

3 0

3 years ago

Identify each part of this chemical equation that describes the burning of methane and oxygen. B (blue box): D (number): E (purp

Allushta [10]

Answer:

The correct answer is -

A (the entire green box): Chemical Equation

B (the blue box): Reactants

C (the arrow): Reacts to Form

D (the number): Coefficient

E (the purple box): Products

Explanation:

The chemical reaction of burning methane and oxygen is as follows;

Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.

2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products

5 0

3 years ago

Estimated speed of the vehicle

SSSSS [86.1K]

Really, Gundy ? ! ?

The formula for the car's speed is given and discussed in the box. The formula is

v = √(2·g·μ·d)

Then they tell you that μ is 0.750 , and then they tell you that d = 52.9 m . Also, everybody knows that 'g' is gravity = 9.8 m/s² .

They also tell us that the mass of the car is 1,000 kg, and they tell us that it took 3.8 seconds to skid to a stop. But we already have all the numbers in the formula without knowing the car's mass or how long it took to stop. The police don't need to weigh the car, and nobody was there to measure how long the car took to stop. All they need is the length of the skid mark, which they can measure, and they'll know how fast the guy was going when he hit the brakes !

Now, can you take the numbers and plug them into the formula ? ! ?

v = √(2·g·μ·d)

v = √( 2 · 9.8 m/s² · 0.75 · 52.9 m)

v = √( 777.63 m²/s²)

v = 27.886 m/s

Rounded to 3 digits, that's 27.9 m/s .

That's about 62.4 mile/hour .

3 0

3 years ago

A force F with arrow applied to an object of mass m1 produces an acceleration of 3.10 m/s2. The same force applied to a second o

Bas_tet [7]

Answer:

(a) The value of the ratio m₁/m₂ is 0.581

(b) the acceleration of the combined masses is 1.139 m/s²

Explanation:

Given;

The acceleration of force applied to M₁, a₁ = 3.10 m/s²

The same force applied to M₂ has acceleration, a₂ = 1.80 m/s²

Let this force = F

According Newton's second law of motion;

F = ma

(a) the value of the ratio m₁/m₂

since the applied force is same in both cases, M₁a₁ = M₂a₂

$\frac{m_1}{m_2} = \frac{a_2}{a_1} \\\\\frac{m_1}{m_2} = \frac{1.8}{3.1} \\\\\frac{m_1}{m_2} = 0.581$

(b) the acceleration of m₁ and m₂ combined as one object under the action force F

F = ma

$a = \frac{F}{M} \\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{F}{0.581m_2 + m_2}\\\\a = \frac{F}{1.581m_2}$

$But, m_2 = \frac{F}{a_2} \\\\a = \frac{F}{1.581m_2} = \frac{F*a_2}{1.581F} \\\\a = \frac{a_2}{1.581} \\\\a = \frac{1.8}{1.581} = 1.139 \ m/s^2$

Therefore, the acceleration of the combined masses is 1.139 m/s²

5 0

3 years ago