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horsena [70]
3 years ago
14

1. Bone has a Young’s modulus of about

Physics
1 answer:
Blababa [14]3 years ago
5 0

#1

As we know that

Y = \frac{stress}{strain}

now plug in all data into this

1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}

strain = 9.33 \times 10^{-3}

now from the formula of strain

strain = \frac{\Delta L}{L}

9.33 \times 10^{-3} = \frac{\Delta L}{0.54}

\Delta L = 5.04 \times 10^{-3} m

\Delta L = 5.04 mm

#2

As we know that

pressure * area = Force

here we know that

Area = 3.53 \times 11.6 = 40.95 m^2

P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa

now force is given as

F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N

#3

As we know that density of water will vary with the height as given below

\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}

here we know that

\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa

B = 2.3 \times 10^9 N/m^2

now density is given as

\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}

\rho = 1185.3 kg/m^3

#4

as we know that pressure changes with depth as per following equation

P = P_o + \rho g h

here we know that

P = 3 P_0

now we will have

3P_0 = P_0 + \rho g h

2P_0 = \rho g h

2(1.01 \times 10^5) = 1025 (9.81)(h)

here we will have

h = 20.1 m

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

mg = \rho_1V_1g + \rho_2V_2g

A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)

46.6 = 43.89 - 922x + 1000x

x = 3.48 cm

so it is 3.48 cm below the interface

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