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NARA [144]
3 years ago
6

You are the technical consultant for an action-adventure film in which a stunt calls for the hero to drop off a 19-m-tall buildi

ng and land on the ground safely at a final vertical speed of 5 m/s. At the edge of the building's roof, there is a 100-kg drum that is wound with a sufficiently long rope (of negligible mass), has a radius of 0.4 m, and is free to rotate about its cylindrical axis with a moment of inertia I0. The script calls for the 76-kg stuntman to tie the rope around his waist and walk off the roof. a) Determine an expression for the stuntman's linear acceleration in terms of his mass m, the drum's radius r, and moment of inertia I0.b) Determine the required value of the stuntman's acceleration if he is to land safely at a speed of 4.00 m/s, and use this value to calculate the moment of inertia of the drum about its axis.c) What is the angular acceleration of the drum?d) How many revolutions does the drum make during the fall?
Physics
1 answer:
xenn [34]3 years ago
7 0

Answer:

D

Explanation:

NOT POSITIVE THO

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A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a
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Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

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\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})

Where:

k - Thermal conductivity, measured in watts per meter-Celsius.

L - Length of the cylinder, measured in meters.

D_{i} - Inner diameter, measured in meters.

D_{o} - Outer diameter, measured in meters.

T_{i} - Temperature at inner surface, measured in Celsius.

T_{o} - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)

If we know that \dot Q = 40.8\,W, L = 0.6\,m, T_{i} = 50\,^{\circ}C, T_{o} = 20\,^{\circ}C, D_{i} = 0.01\,m and D_{o} = 0.04\,m, the thermal conductivity of the biomaterial is:

k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)

k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

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Read 2 more answers
40 POINTS!!!
Feliz [49]

1.) B.

2.) A.

3.) A.

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