Which of the following is not an example of energy transfer?

iragen [17]

The Moment of Inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Let suppose that the Disk is a Rigid Body whose mass is uniformly distributed. The Moment of Inertia of the element is equal to the Moment of Inertia of the entire Disk minus the Moment of Inertia of the Hole, that is to say:

$I = I_{D} - I_{H}$ (1)

Where:

$I_{D}$ - Moment of inertia of the Disk.

$I_{H}$ - Moment of inertia of the Hole.

Then, this formula is expanded as follows:

$I = \frac{1}{2}\cdot M\cdot R^{2} - \frac{1}{2}\cdot m\cdot \left(\frac{1}{2}\cdot R^{2} \right)$ (1b)

Dimensionally speaking, Mass is directly proportional to the square of the Radius, then we derive the following expression for the Mass removed by the Hole ( $m$ ):

$\frac{m}{M} = \frac{R^{2}}{4\cdot R^{2}}$

$m = \frac{1}{2}\cdot M$

And the resulting equation is:

$I = \frac{1}{2}\cdot M\cdot R^{2} -\frac{1}{2}\cdot \left(\frac{1}{4}\cdot M \right) \cdot \left(\frac{1}{4}\cdot R^{2} \right)$

$I = \frac{1}{2} \cdot M\cdot R^{2} - \frac{1}{32}\cdot M\cdot R^{2}$

$I = \frac{15}{32}\cdot M\cdot R^{2}$

The moment of inertia of the Disc is represented by $I = \frac{15}{32}\cdot M\cdot R^{2}$ . (Correct answer: A)

Please see this question related to Moments of Inertia: brainly.com/question/15246709

5 0

2 years ago

The amplitude of a simple harmonic oscillator will be doubled by:a) doubling only the initial speedb) doubling the initial displ

OLEGan [10]

Answer:

When both initial speed and initial displacement is doubled then amplitude will be doubled.

Explanation:

Given that :- Amplitude of simple harmonic Oscillator is doubled.

So,

Formula of Simple harmonic oscillator is $X=A\sin\ (2\pi ft +\phi)$ ...........(1)

Where X = Position in (m,cm,km.....)

A = Amplitude in (m,cm,km.....)

F = Frequency in (Hz)

T = Time in (sec.)

Ф = Phase in (rad)

For initial displacement taking t=0 we get,

Initial displacement = $A\sin(\phi)$ .................(2)

Now taking equation (1) and differentiating it w.r.t to (t) we get

$\frac{dx}{dt} = 2\pi fA\cos\ (2\pi ft+\phi)$

$V= 2\pi fA\cos\ (2\pi ft+\phi)$

taking t=0 for initial speed then we get,

Initial speed = $2\pi fA\cos\phi$ ...............(3)

observing equation (2) & (3) that the initial displacement and initial speed depends on the Amplitude of the Oscillator.

Hence,

when both initial speed and displacement is doubled then amplitude will be doubled.

4 0

3 years ago

Name any two liquid which are found in our human eyes

Romashka [77]

Aqueous humor and vitreous humor are the liquids present in the human eye.

Hope it helped you... pls mark brainliest

8 0

2 years ago

For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro

viktelen [127]

Answer:

$V_d$ = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, $A = \frac{\pi d^2}{4}$ = $A = \frac{\pi 0.625^2}{4}$

Now,

The current density, J is given as

$J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}$ = 2444619.925 A/mm²

now, the electron density, $n = \frac{\rho}{M}N_A$

where,

$N_A$ =Avogadro's Number

$n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3$

Now,

the drift velocity, $V_d$

$V_d=\frac{J}{ne}$

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

$V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e}$ = 1.75 × 10⁻⁴ m/s

4 0

2 years ago

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What is the correct chemical name for C7H8?

Nesterboy [21]

Answer:

Cycloheptatriene

Explanation:

4 0

2 years ago

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