Answer:
The moment of inertia is
Explanation:
The moment of inertia is equal:
If r is
and
A positively charged ebonite rod is brought close to a small ball. The rod does not touch the ball, which is made from a conduct
1 answer:
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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=14$
X=14:100
For more information on fraction
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?
<u>Answer:</u>
Work done done by a system on its environment is W=40 Joules
<u>Explanation:</u>
<u>Given:</u>
Ideal gas is maintained at a constant pressure
Change in Volume
<u>To find:</u>
Work done by a system during an Isobaric Process
<u>Solution:</u>
In a Isobaric Process, constant pressure is maintained,
According to the formula,
work done
Substitute the values of pressure and change in volume in the above formula,
<u>Result:</u>
Work done done by a system W= -40 Joules