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geniusboy [140]

2 years ago

9

A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.60 m and a mass of 0.44 kg. The rod has a

length of 1.28 m and a mass of 0.48 kg. The rod is placed on a fulcrum (pivot) at X = 0.46 m from the left end of the rod. Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum.

1 answer:

siniylev [52]2 years ago

3 0

Answer:Moment of inertia is the property of a body due to which it resists angular acceleration. It is the sum of the products of the mass of each particle in the body with the square of its distance from the axis of rotation.

Explanation: I = m*r*r

Given mass of sphere = 0.4kg

Radius of sphere = 0.6m

Moment of Inertia of sphere = (2/3)*0.4* 0.6

Inertia of sphere = 0.096 kgm2

Given mass of rod = 0.48kg

Length of rod = 1.28m

Inertia of rod = (m*L*L)/3

I = (0.48 * 1.28 * 1.28)/3

I = 0.218 kgm2

Total Moment of inertia I = 0.096 + 0.218

I = 0.314 kgm2

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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

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Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the hear

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From image B attached, we can calculate the current flowing through the capacitors.

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Since V=IR; I = V/R = 5/500 = 0.01 A

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A 4kg bird has 8 joules of kinetic energy, how fast is it flying?

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I believe the answer is 2m/s

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B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

$F=mg$

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

$a_y = g = -9.8 m/s^2$

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C) 7.6 m/s

The x-component of the shot's velocity is given by:

$v_x = v_0 cos \theta$

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Substituting into the equation, we find

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D) 9.3 m/s

The y-component of the shot's velocity is given by:

$v_y = v_0 sin \theta$

where

$v_0 = 12.0 m/s$ is the initial velocity

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Substituting into the equation, we find

$v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s$

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

$v_x = 7.6 m/s$

F) -11.1 m/s

The y-component of the velocity at time t is given by:

$v_y(t) = v_y + at$

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$v_y = 9.3 m/s$ is the initial y-velocity

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Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

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