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Andrei [34K]
2 years ago
15

Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m

ass, but has hexagonal sections where material has been removed. The attached masses are released from rest and allowed to fall a height h.Which of the following statements about their angular accelerations is true? a. The angular acceleration of the two flywheels is different but it is impossible to tell which is greater. b. The angular acceleration of flywheel A is greater The angular acceleration of flywheel B is greater. c. Not enough information is provided to determine. d. The angular accelerations of the two flywheels are equal.
Physics
1 answer:
inysia [295]2 years ago
4 0

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

      τ = I α

     T r = I α

     T gR = I α

     Alf = T R / I (1)

     T = α I / R

Now let's use Newton's second law in the mass that descends

     W- T = m a

     a = (m g -T) / m

The two accelerations need related

     a = R α

    α = a / R

    a = (m g - α I / R) / m

    R α = g - α I /m R

    α (R + I / mR) = g

    α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

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Answer:

We need to multiply 12 to each term to eliminate fractions.

Explanation:

Given expression:

6-\frac{3}{4}x+\frac{1}{3}=\frac{1}{2}x+5

To eliminate the fraction we need to multiply each term by least common multiple of the denominators of the fraction.

The denominators in the above expressions are:

4, 3 and 2

The multiples of each can be listed below.

2⇒ 2,4,6,8,10,<u>12</u>,14,16.....

3⇒ 3,6,9,<u>12</u>,15,18

4⇒ 4,8,<u>12</u>.......

From the list of the multiples stated, we can see the least common multiple is 12.

So we will multiply each term by 12.

Multiplying 12 to both sides.

12(6-\frac{3}{4}x+\frac{1}{3})=12(\frac{1}{2}x+5)

Using distribution,

72-9x+4=6x+60

Thus we successfully eliminated the fractions.

5 0
2 years ago
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what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t
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Recall this equation for a device in a direct current circuit:
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Technician A says that the use of some RTV sealants to seal components on an engine can damage the oxygen sensor. Technician B s
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RTV ( Room Temperature Vulcanising) sealants are silicon products used for the sealing purpose of the engine component. As the name implies once exposed to air, it solidifies. However, it has a negative effect on the oxygen sensor of the engine components by makes a coating over the sensors with the silicone film and, after a time, will affect the operation performed by them. Therefore Technician A is correct.

Technician B is equally correct in that an engine need to be lubricated with oil to help cool the moving part while on. But in order to ensure that oil reaches all movable parts a common method know as heavy-duty reversible spin drills is applied. This method regulates the flow of oil using the engine.

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2 years ago
A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, \lambda_1=8\times 10^{-6}\ C

linear charge density of the other charge distribution, \lambda_2=-6\times 10^{-6}\ C

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

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\lambda= linear charge density

r = radial distance from the center of wire

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Therefore,

E_1=E_2

\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}

\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}

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