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Viktor [21]
3 years ago
12

Larger animals have sturdier bones than smaller animals. A mouse's skeleton is only a few percent of its body weight, compared t

o 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 10−4 m2, a typical value.)
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?
Physics
1 answer:
Rama09 [41]3 years ago
7 0

Answer:

a_s=4.8\times  10^{-2}~m^2

Explanation:

Given:

cross sectional area of the bone, a=4.8 \times 10^{-4} ~m^2

factor of up-scaling the dimensions, s=10

Since we need to find the upscaled area having two degrees of the dimension therefore the scaling factor gets squared for the area being it in 2-dimensions.

The scaled up area is:

a_s=a\times s^2

a_s=[4.8 \times 10^{-4}]\times 10^2

a_s=4.8\times  10^{-2}~m^2

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Bad White [126]

Answer: 5.96m/s

Explanation:

Given the following :

Mass of car (m) = 1500kg

Velocity (V) = 5.25m/s

Forward force of engine = 1250N

Diatance moved = 4.8m

Final Velocity =?

Final kinetic energy = Initial kinetic energy + work done by engine

Initial kinetic energy = 0.5 × mass × velocity^2

Initial kinetic energy = 0.5 × 1500 × 5.25^2

Initial kinetic energy = 20671.875 J

Work done by engine = Force × distance

Work done by engine = 1250 × 4.8 = 6000J

Final kinetic energy = (20671.875 + 6000) J

= 26671.875 J

From kinetic energy = 0.5mv^2

26671.875 = 1/2 × 1500 × v^2

53343.75 = 1500v^2

v^2 = 35.5625

v = sqrt(35.5625)

v = 5.96m/s

3 0
3 years ago
1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol
torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0
3 years ago
A hole is drilled in a metal plate. When the metal is raised to a higher temperature, what happens to the diameter of the hole?
zysi [14]

Answer:

The diameter of the hole increases

Explanation:

Metals expand and contract with temperature. Whenever metal is heated, it usually expands in relation to its thermal expansivity. This expansion leads to a slight increase in surface area.

Once the surface area of the metal changes, this means that the dimensions of the whole metal surface changed. As a result, the diameter of the hole drilled in the metal plate will change also. In our case, the diameter of the hole will increase.

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A 12 kg box is at rest on your kitchen counter, which your cat is pawing at with a horizontal force of 40 N. If the coefficient
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Answer; I think it's False.

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