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Igoryamba
2 years ago
15

If the scaled-up man now stands on one leg, what fraction of the tensile strength is the stress on the femur?.

Physics
1 answer:
Firdavs [7]2 years ago
8 0

The new cross sectional area and tensile stress is mathematically given as

A=4.8*10^{-4}m^2

X=14:100

<h3>Cross sectional area and stress on the femur</h3>

Question Parameters:

To see why this must be so, recall, that the stress on the femur for a man standing on one leg is 1.4%

we scale this man up by a factor of 10 in all dimensions

that a 70 kg person has a femur with a cross-section area

a)

Generally the New cross sectional area  is mathematically given as

100A=100*4.8*10^{-4}

A=4.8*10^{-4}m^2

b)

From the initial statement we see that  the fraction of the tensile strength is the stress on the femur  and will be

X=\frac{14}{100}

X=14$

X=14:100

For more information on fraction

brainly.com/question/1301963

Complete Question

Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared to 16% for an elephant. To see why this must be so, recall, that the stress on the femur for a man standing on one leg is 1.4% of the bone's tensile strength. Suppose we scale this man up by a factor of 10 in all dimensions, keeping the same body proportions. (Assume that a 70 kg person has a femur with a cross-section area (of the cortical bone) of 4.8 x 10-4m², a typical value.)

Part A

Both the inside and outside diameter of the femur, the region of cortical bone, will increase by a factor of 10. What will be the new cross-section area? Express your answer using two significant figures.

Part B

If the scaled-up man now stands on one leg, what fraction of the tensile strength is the stress on the femur?

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a. 192 m/s

b. -17,760 kPa

Explanation:

First let's write the flow rate of the liquid, using the following equation:

Q = A*v

Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:

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v2 = 0.02^2 * 12 / 0.005^2

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b.

To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)

P1 + d1*v1^2/2 = P2 + d1*v2^2/2

Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)

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