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Contact [7]
3 years ago
7

You own a yacht which is 14.5 meters long. It is motoring down a canal at 10.6 m/s. Its bow (the front of the boat) is just abou

t to begin passing underneath a bridge that is 30.0 m across. How much time is required until its stern (the end of the boat) is no longer under the bridge?
Physics
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

4.198 seconds

Explanation:

The total length of the yacht is 30m and the width of the bridge is 14.5m.

The speed of the yacht is 10.6m/s.

We can assume the current of the canal has a speed of 0m/s.

To calculate the answer we will break it into two parts

1) The time it takes for the bow to travel from the start of the bridge to the end

<em>Formula Speed = Distance / Time</em>

Speed = 10.6m/s

Distance = 30m

Time = T1

T1 = 2.8301s

2) The time it takes for the stern to cross the end of the bridge after the bow

<em />

<em>Formula Speed = Distance / Time</em>

<em />

Speed = 10.6m/s

Distance = 14.5m

Time = T2

T2 = 1.3679s

3) The Total Time is

Total = T1 + T2

Total = 4.198s

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Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate
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Answer:

\epsilon_2=0.098

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Temperature at base T_b=400k

Generally the equation for Area of base surface is mathematically given by

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 A_b=\frac{\pi}{4}0.3^2

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Generally the equation for Area of Hemispherical dome is mathematically given by

 A_h=\frac{\pi}{2}d^2

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 F_{11}+F_{12}=1

 F_{11}=0

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 F_{12}=1

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Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

 Q_{21}=-Q_{12}

 Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }

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 \epsilon_2 \approx 0.1

Therefore  the emissivity of the dome is

 \epsilon_2=0.098

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