PHYSICS COLLEGE

A 133 kg horizontal platform is a uniform disk of radius 1.95 m and can rotate about the vertical axis through its center. A 62.7 kg person stands on the platform at a distance of 1.19 m from the center, and a 28.5 kg dog sits on the platform near the person 1.45 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.

Answers

Answer 1

Answer:

Answer:

The moment of inertia of the system is $I = 400.5 \ kg \cdot m^2$

Explanation:

From the question we are told that

The mass of the platform is $m = 133\ kg$

The radius of the platform is r = 1.95 m

The mass of the person is $m_p = 62.7 \ kg$

The position of the person from the center is $d = 1.19 \ m$

The mass of the dog is $m_D = 28.5 \ kg$

The position of the dog from the center is $D = 1.45 \ m$

Generally the moment of inertia of the platform with respect to its axis is mathematically represented as

$I_p = (m r^2)/(2)$

The moment of inertia of the person with respect to the axis is mathematically represented as

$I_z = m_p* d^2$

The moment of inertia of the dog with respect to the axis is mathematically represented as

$I_D = m_d * D^2$

So the moment of inertia of the system about the axis is mathematically evaluated as

$I = I_p + I_z + I_D$

=> $I = (mr^2)/(2) + m_p * d^2 + m_d * D^2$

substituting values

$I = ((133) * (1.95)^2)/(2) + (62.7) * (1.19)^2 + (28.5) * (1.45)^2$

$I = 400.5 \ kg \cdot m^2$

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Answers

Answer:

The flux through the surface of the cube is $2.314\ Nm^(2)/C$

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = $8.0* 10^(- 2)\ m$

Volume Charge density, $\rho_(v) = 40 nC/m^(3) = 40* {- 9}\ C/m^(3)$

Now,

To calculate the electric flux:

$\phi = (q)/(\epsilon_(o))$ (1)

where

$\phi$ = electric flux

$\epsilon_(o) = 8.85* 10^(- 12)\ F/m$ = permittivity of free space

Volume Charge density for the given case is given by the formula:

$\rho_(v) = (Total\ charge, q)/(Volume of cube, V)$ (2)

Volume of cube, $V = a^(3)$

Thus

$V = (8.0* 10^(- 2))^(3) = 5.12* 10^(- 4)\ m^(3)$

Thus from eqn (2), the total charge is given by:

$q = \rho_(v)V = 40* {- 9}* 5.12* 10^(- 4)$

$q = 2.048* 10^(-11)\ F = 20.48\ pF$

Now, substitute the value of 'q' in eqn (1):

$\phi = (2.048* 10^(-11))/(8.85* 10^(- 12)) = 2.314\ Nm^(2)/C$

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Answers

Answer:

The answer is below

Explanation:

The speed of the boat in still water is perpendicular to the speed of the water flow. Therefore the speed relative to the ground (V), the speed of flow and the speed of the boat in still water form a right angled triangle. Hence the speed relative to the ground is given as:

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Answers

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Answers

Answer:

The force must be applied on the axis of rotation

Explanation:

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Answers

Answer:

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Explanation:

From the question we are told that

The primary voltage is $V_p = 120 \ V$

The secondary voltage is $V_s = 6 \ V$

Generally from the transformer equation we have that

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Answers

Answer:b

Explanation:

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