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3 years ago

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A uniform steel rod of length 0.9 m and mass 3.8 kg has two point masses of 2.3 kg each at the two ends. Calculate the moment of

inertia of the system about an axis perpendicular to the rod, and passing through its center.

1 answer:

polet [3.4K]3 years ago

3 0

Answer: $2.4705\ kg.m^2$

Explanation:

Given

length of the rod is L=0.9 m

Mass of the rod m=3.8 kg

Point masses has mass of m=2.3 kg

Moment of Inertia of the rod about the center is

$\Rightarrow I_o=\dfrac{1}{12}ML^2$

Moment of inertia of combined system is the sum of rod and two point masses.

$\Rightarrow I=I_o+2mr^2$

$\Rightarrow I=\dfrac{1}{12}3.8\times 0.9^2+2\times 2.3\times \left(\dfrac{0.9}{2}\right)^2\\\\\Rightarrow I=1.539+0.9315\\\Rightarrow I=2.4705\ kg-m^2$

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Answer:

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Explanation:

Given:

time 't' = 2.9s

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distance 'd'=0.7m

The time period 't' is the time b/w two successive waves. Therefore, the time it takes from the boat to travel from its highest point to its lowest is a half period.

So, T = 2 x 2.9 => 5.8 s

As we know that frequency is the reciprocal of time period, we have

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In order to find how fast are the waves traveling, the velocity is given by

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V= 0.2 x 5.5 =>1.1m/s

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A bird flies 3.7 meters in 46 seconds, what is its speed?

victus00 [196]

Answer:

Speed is 0.08 m/s.

Explanation:

Given the distance that the bird flies = 3.7 meters

The time is taken by the bird to fly the 3.7 meters = 46 seconds

We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.

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3 years ago

A bus accelerates from 5.75 m/s at a rate of 1.25 m/s/s for 3.50

marysya [2.9K]

Answer:

10.125 meters?

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3 years ago

A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.

34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated ( $Q$ ), measured in joules, by the cylindrical resistor is the power multiplied by operation time ( $\Delta t$ ), measured in hours. That is:

$Q = \dot Q \cdot \Delta t$ (1)

If we know that $\dot Q = 1.2\,W$ and $\Delta t = 86400\,s$ , then the amount of heat dissipated by the resistor is:

$Q = (1.2\,W)\cdot (86400\,s)$

$Q = 103680\,J$

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux ( $Q'$ ), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder ( $A$ ), measured in square meters:

$Q' = \frac{\dot Q}{A}$ (2)

$Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h }$ (3)

Where:

$D$ - Diameter, measured in meters.

$h$ - Length, measured in meters.

If we know that $\dot Q = 1.2\,W$ , $D = 4\times 10^{-3}\,m$ and $h = 2\times 10^{-2}\,m$ , the heat flux of the resistor is:

$Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }$

$Q' \approx 4340.589\,\frac{W}{m^{2}}$

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces ( $r$ ), no unit, is the ratio of the top and bottom surfaces to total surface:

$r = \frac{\frac{\pi}{2}\cdot D^{2}}{A}$ (3)

If we know that $A \approx 2.765\times 10^{-4}\,m^{2}$ and $D = 4\times 10^{-3}\,m$ , then the fraction is:

$r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}$

$r = 0.045$

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

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3 years ago