Genes do not affect more than our physical characteristics True False
1 answer:
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Answer:
f = - 0.143 m = - 14.3 cm
Explanation:
First, we will calculate the power of the combination of lenses:
P = P₁ + P₂
where,
P = Power of Combination = ?
P₁ = Power of first lens = - 5D
P₂ = Power of second lens = - 2D
Therefore,
P = - 5D - 2D
P = - 7D
Now, the focal length can be given as:
<u>f = - 0.143 m = - 14.3 cm</u>
Negative focal length indicates that combination will act as diverging lens.
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?
Answer:4.08 N
Explanation:
Given data
superball dropped from a height of 10 cm
Mass of ball=0.05kg
time of contact=34.3
s
Now we know impulse ==Change in momentum
=
and velocity at the bottom is given by
v=
=
=4.081N