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1. Bone has a Young’s modulus of about
1.8 × 1010 Pa . Under compression, it can
withstand a stress of about 1.68 × 108 Pa before breaking.
Assume that a femur (thigh bone) is 0.54 m
long, and calculate the amount of compression
this bone can withstand before breaking.
Answer in units of mm.

2. If a 1-megaton nuclear weapon is exploded
at ground level, the peak overpressure (that
is, the pressure increase above normal atmospheric pressure) will be 0.2 atm at a distance
of 6 km.
What force due to such an explosion will be
exerted on the side of a house with dimensions
3.53 m × 11.6 m?
Answer in units of N

3. Find the density of seawater at a depth where
the pressure is 2600 atm if the density at the
surface is 1050 kg/m3
. Seawater has a bulk
modulus of 2.3 × 109 N/m
2
.
Answer in units of kg/m3
.

4. Calculate the depth in the ocean at which the
pressure is three times atmospheric pressure.
The acceleration of gravity is 9.81 m/s
2
and
the density of sea water is 1025 kg/m3
.
Answer in units of m.

4. Oil having a density of 922 kg/m
3
floats on
water. A rectangular block of wood 4.76 cm
high and with a density of 979 kg/m3 floats
partly in the oil and partly in the water. The
oil completely covers the block.
How far below the interface between the
two liquids is the bottom of the block?
Answer in units of m.

5. A fireman standing on a 9.2 m high ladderoperates a water hose with a round nozzle of
diameter 3.3 inch. The lower end of the hose
(9.2 m below the nozzle) is connected to the
pump outlet of diameter 4.59 inch. The gauge
pressure of the water at the pump is
P
(gauge)
pump = P
(abs)
pump − Patm
= 47.2 PSI = 325.433 kPa .
Calculate the speed of the water jet emerging from the nozzle. Assume that water is an
incompressible liquid of density 1000 kg/m3
and negligible viscosity. The acceleration of
gravity is 9.8 m/s
2
.
Answer in units of m/s.

6. A horizontal pipe of diameter 0.806 m has a
smooth constriction to a section of diameter
0.4836 m . The density of oil flowing in the
pipe is 821 kg/m3
.
If the pressure in the pipe is 8350 N/m
2
and in the constricted section is 6262.5 N/m2
,
what is the rate at which oil is flowing?
Answer in units of m3/s

7. An object weighing 305 N in air is immersed
in water after being tied to a string connected
to a balance. The scale now reads 274 N .
Immersed in oil, the object appears to weigh
278 N .
Find the density of the object.
Answer in units of kg/m
Find the density of the oil.
Answer in units of kg/m3.



This is very important please if you can solve before tomorrow thx

Respuesta :

#1

As we know that

[tex]Y = \frac{stress}{strain}[/tex]

now plug in all data into this

[tex]1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}[/tex]

[tex]strain = 9.33 \times 10^{-3}[/tex]

now from the formula of strain

[tex]strain = \frac{\Delta L}{L}[/tex]

[tex]9.33 \times 10^{-3} = \frac{\Delta L}{0.54}[/tex]

[tex]\Delta L = 5.04 \times 10^{-3} m[/tex]

[tex]\Delta L = 5.04 mm[/tex]

#2

As we know that

pressure * area = Force

here we know that

[tex]Area = 3.53 \times 11.6 = 40.95 m^2[/tex]

[tex]P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa[/tex]

now force is given as

[tex]F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N[/tex]

#3

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

[tex]\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}[/tex]

here we know that

[tex]\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa[/tex]

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

now density is given as

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

[tex]\rho = 1185.3 kg/m^3[/tex]

#4

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

[tex]P = P_o + \rho g h[/tex]

here we know that

[tex]P = 3 P_0[/tex]

now we will have

[tex]3P_0 = P_0 + \rho g h[/tex]

[tex]2P_0 = \rho g h[/tex]

[tex]2(1.01 \times 10^5) = 1025 (9.81)(h)[/tex]

here we will have

[tex]h = 20.1 m[/tex]

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

[tex]mg = \rho_1V_1g + \rho_2V_2g[/tex]

[tex]A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)[/tex]

[tex]46.6 = 43.89 - 922x + 1000x[/tex]

[tex]x = 3.48 cm[/tex]

so it is 3.48 cm below the interface

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