All Solutions
Page 446: Assessment
b) The air in refrigerator will always be at atmospheric pressure since it we are cooling it slowly so the only correct explanation is A.
– The mass of the person $m_{text{p}} = 79text{ kg}$
– The mass of the chair $m_{text{c}} = 3.7text{ kg}$
– The diameter of each leg of the chair $d = 1.3text{ cm}$
The total weight of the person sitting on the chair is balanced by the force between the floor and the chair legs so we can write:
$$
begin{gather*}
m_{text{total}} cdot g = A_{text{total}} cdot P \
left(m_{text{p} + m_{text{c}}} right)g = 4 cdot left(frac{d}{2} right)^2 pi , P
end{gather*}
$$
Where we have taken into account that there are four legs.
Rearranging for $P$ we have:
$$
begin{align*}
P = frac{(m_{text{p}} + m_{text{c}} )g }{d^2 , pi }
end{align*}
$$
Plugging in the values we get:
$$
begin{align*}
P = frac{(79text{ kg} + 3.7text{ kg} ) cdot 9.81 ; frac{text{m}}{text{s}^2} }{(0.013text{ m})^2 , pi }
= 1.53 cdot 10^{6}text{ Pa}
end{align*}
$$
$$
p_1 = frac{F}{r_1^2pi}quad p_2 = frac{F}{r_2^2pi}
$$
which yield
$$
frac{p_2}{p_1} = frac{r_1^2}{r_2^2} = 0.23
$$
which is the reduction for $77%$,
$$
p=p_{gauge} + p_{atm} = 169text{ kPa}.
$$
$$
p_1V=n_1RTquad p_2V=n_2RT.
$$
Subtracting these equations after elementary transformations we get
$$
Delta n = n_2-n_1 = frac{(p_2-p_1)V}{RT} = 0.3text{ mol}.
$$
$$
pV=nRT=frac{m}{M}RT
$$
where $M=4text{ g/mol}$ is the molar mass of helium. This gives
$$
m=frac{pVM}{RT} =1327text{ kg}.
$$
– The mass of the car $m_{text{c}} = 1420text{ kg}$
– The gauge pressure of each tire $P_{text{gauge}} = 351text{ kPa}$
begin{enumerate}[a)]
item
The total weight of the car is balanced by the force between the ground and the tires so we can write:
begin{align*}
m_{text{total}} cdot g &= A_{text{total}} cdot P \
m_{text{c}} cdot g &= 4 cdot A cdot P_{text{gauge}}
end{align*}
Where we have taken into account that there are four tires. \\
Rearranging for $A$ we have:
begin{align*}
A = frac{m_{text{c}} , g }{4 , P_{text{gauge}} }
end{align*}
Plugging in the values we get:
begin{align*}
A = frac{1420text{ kg} cdot 9.81 ; frac{text{m}}{text{s}^2} }{4 cdot 351 cdot 10^{3}text{ Pa}}
= 9.92 cdot 10^{-3}text{ m}^2
approx 0.01text{ m}^2
end{align*}
item
We see that the product of the gauge pressure and the contact area equals the weight of the car, hence:
begin{align*}
A cdot P_{text{gauge}} = text{const}
end{align*}
So when the pressure is increased the area decreases.
end{enumerate}
item
The area of each tire is $A = 0.01text{ m}^2$
item
When the pressure is increased the area decreases.
end{enumerate}
$$
pV=nRT.
$$
If $p$ is kept constant we easily conclude that the ratio $V/T$ is also constant so we can write
$$
frac{V_1}{T_1}=frac{V_2}{T_2}Rightarrow frac{Ah_1}{T_1}=frac{Ah_2}{T_2}Rightarrowfrac{h_1}{T_1}=frac{h_2}{T_2}.
$$
Cross-multiplying the last equation we get
$$
T_2 = frac{h_2}{h_1}T_1 =348text{ K}.
$$
On the Moon the force of gravity is about 6 times weaker than on the Earth. This means that if we are drinking out of a can (whose pressure is high) it will be easier to sip that it is on the Earth. Now how that can didn’t explode because of the vacuum around it is another question entirely.
Because of that vacuum drinking out of a glass or other open container would be impossible, it would require the creation of negative pressure in our mouth.
$$
F=p_{atm}A = p_{atm} dpi h =2.5text{ kN}.
$$
$$
p_{atm}= rho_{Hg}gh = 98text{ kPa}.
$$
$$
p=p_{atm} + rho g h
$$
which yields
$$
h=frac{p-p_{atm}}{rho g}=1.9text{ m}.
$$
item
The pressure at a given depth $h$ in the rooftop water storage tower can by found by adding the hydrostatic pressure for that depth, and the pressure at the top of the tower. \\
At the top of the tower the pressure is $P_{text{atm}}$ so we write:
begin{align*}
P = P_{text{atm}} + rho cdot g cdot h
end{align*}
Plugging in the values we get:
begin{align*}
P &= 101.3text{ kPa} + 1000 ; frac{text{kg}}{text{m}^3} cdot 9.81 ; frac{text{m}}{text{s}^2} cdot 4.5text{ m} \
P&=145.5text{ kPa}
end{align*}
item
As we have already explained the pressure in the water tower increases with depth, so the bottom part of the tower requires more support to prevent instabilities (bursting). Hence, more metal bands are needed.
end{enumerate}
item
$P = 145.5text{ kPa}$
item
The pressure increases with depth in the water tower, hence more metal bands are needed to stabilize the tower and prevent bursting.
end{enumerate}
$$
P=rho g h
$$
which gives for the depth
$$
h= frac{P}{rho g} = 990text{ m}.
$$
NOTE:
$$
P=10text{ }frac{text{N}}{text{mm}^2} = 10timesfrac{N}{(0.001text{ m})^2} = 10^7text{ Pa}.
$$
b) The safe depth is greater since fresh water has lower density and thus exerts less pressure at the same depth.
$$
p-p_{atm} = rho g h
$$
which yields
$$
h=frac{p-p_{atm}}{rho g} = 77text{ cm}.
$$
b) I we use a less dense fluid the suspension point has to be higher since it would exert less additional static pressure that more dense fluid (note that the pressure at a fixed depth $d$ is $rho g d$ i.e. it is directly proportional to $rho$).
$$
rhotext{ height}timestext{width}timestext{depth }g = (m+m_m)g.
$$
Solving this for $m$ we get
$$
m = rhotext{ height}timestext{width}timestext{depth }- m_m = 212.3text{ kg}.
$$
$$
p_{atm} +frac{m_1 g}{r_1^2pi} =p_{atm} + rho g h + frac{m_2 g}{r_2^2pi}.
$$
Solving for $h$ we get
$$
h = frac{m_1r_2^2-m_2r_1^2}{r_1^2r_2^2rhopi} = 1.2text{ m}.
$$
As implied by the Equation of Continuity as the speed of the fluid flow increases, its cross sectional area will decrease since:
$$
begin{equation*}
V cdot A = text{const}
end{equation*}
$$
So the sheet of water will be thinner at the bottom of the waterfall since its speed is greater.
This force is proportional to the speed with which the airplane moves.
When wind is blowing the relative speed of the plane to the wind is relevant. To maximize the relative speed the airplane should be moving into the wind.
$$
v_{hose}A_{hose} = v_{nozzle} A_{nozzle}.
$$
We easily obtain
$$
v_{nozzle} = v_{hose}frac{A_{hose}}{A_{nozzle}} = v_{hose}frac{(d_{hose}/2)^2pi}{(d_{nozzle}/2)^2pi} = v_{hose}left( frac{d_{hose}}{d_{nozzle}}right)^2 = 8.22text{ m/s}.
$$
$$
v_{hose}A_{hose} = v_{nozzle} A_{nozzle}.
$$
We easily obtain
$$
v_{nozzle} = v_{hose}frac{A_{hose}}{A_{nozzle}} = v_{hose}frac{(d_{hose}/2)^2pi}{(d_{nozzle}/2)^2pi} = v_{hose}left( frac{d_{hose}}{d_{nozzle}}right)^2.
$$
This finally yields
$$
d_{nozzle} = d_{hose}sqrt{frac{v_{hose}}{v_{nozzle}}} =0.66text{ cm}.
$$
$$
v_{normal}A_{normal} = v_{plaque} A_{plaque}.
$$
We easily obtain
$$
v_{plaque} = v_{normal}frac{A_{normal}}{A_{plaque}} = v_{normal}frac{(d_{normal}/2)^2pi}{(d_{plaque}/2)^2pi} = v_{normal}left( frac{d_{normal}}{d_{plaque}}right)^2 =32text{ cm/s}.
$$
$$
v_{wide}A_{wide} = v_{narrow}A_{narrow}Rightarrow v_{narrow} = v_{wide}frac{A_{wide}}{A_{narrow}} = 4v_{wide} = 6.4text{ m/s}.
$$
b) Since the speed of water increases (and so its’ kinetic energy with no change in height and thus in the potential energy) the pressure decreases according to Bernoulli’s principle so the pressure is lower than $110text{ kPa}.$
Hence the object will not return to its initial shape and size after the force that acts on it stops, as was the case for elastic deformations.
– The elongation of the spring $x = 9.7 text{ cm}$
– The spring constant of the spring $k = 61 ; frac{text{N}}{text{m}}$
When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.
We will solve the problem by applying Hooke’s Law :
$$
begin{equation*}
F = k , x
end{equation*}
$$
Plugging in the numbers we get:
$$
begin{align*}
F = 61 ; frac{text{N}}{text{m}} cdot 0.097text{ m} approx 5.9text{ N}
end{align*}
$$
– The elongation of the spring $x = 12 text{ cm}$
– The spring constant of the spring $k = 57 ; frac{text{N}}{text{m}}$
When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.
We will solve the problem by applying Hooke’s Law :
$$
begin{equation*}
F = k , x
end{equation*}
$$
Plugging in the numbers we get:
$$
begin{align*}
F = 57 ; frac{text{N}}{text{m}} cdot 0.12text{ m} approx 6.8text{ N}
end{align*}
$$
– The force acting on a spring $F = 5.2 text{ N}$
– The elongation of the spring $x = 11 text{ cm}$
When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.
We will solve the problem by applying Hooke’s Law :
$$
begin{equation*}
F = k , x
end{equation*}
$$
Rearranging for $k$ we have:
$$
begin{equation*}
k = frac{F}{x}
end{equation*}
$$
Plugging in the numbers we get:
$$
begin{align*}
k = frac{5.2text{ N}}{0.11text{ m}} approx 47.3 ; frac{text{N}}{text{m}}
end{align*}
$$
– The force acting on a spring $F = 51 text{ N}$
– The spring constant of the spring $k = 38 ; frac{text{kN}}{text{m}}$
When a spring is compressed or pulled by a given force $F$ its elongation $x$ is proportional to that force, and the coefficient is called the spring constant.
We will solve the problem by applying Hooke’s Law :
$$
begin{equation*}
F = k , x
end{equation*}
$$
Rearranging for $x$ we have:
$$
begin{equation*}
x = frac{F}{k}
end{equation*}
$$
Plugging in the numbers we get:
$$
begin{align*}
x = frac{51 text{ N}}{38,000 ; frac{text{N}}{text{m}}} approx 0.00134text{ m}
end{align*}
$$
$$
mg=kx
$$
which gives
$$
k=frac{mg}{x} = 1635text{ N/m}.
$$
$$
mg=kx.
$$
This gives
$$
m=frac{kx}{g} = 2.5times 10^{-4}text{ kg}.
$$
$$
T=kDelta l
$$
where $T$ is tension. This yields
$$
k=frac{T}{Delta l} =frac{380text{ N}}{0.13times10^{-2}text{ m}} = 292text{ kN/m}.
$$
Obviously the volume of water is greatest in container C, since there is no other object in it.
For the remaining two containers, the volume of water in them is reduced by the amount pushed out by the wooden block.
When the lead weight is placed on the wooden block in container B it sinks even further into the water than the one in container A.
Hence the volume below the water level occupied by the block is greater for container B than it is for A.
We finally conclude that the volumes of water are ranked as follows:
$$
begin{align*}
V_{text{C}} > V_{text{A}} > V_{text{B}}
end{align*}
$$
In order of increasing weight of the water, the containers are ranked as follows:
$$
begin{align*}
B < A < C
end{align*}
$$
$$
begin{align*}
B < A < C
end{align*}
$$
$$
mg = kx.
$$
Now we have to calculate the mass of the block. From the definition of density $rho = m/V$ we get $m=rho V$. Putting this to the first equation we get
$$
rho V g = kx
$$
which yields for the elongation
$$
x=frac{rho V g}{k} =15.5text{ cm}.
$$
$$
F=p_{over}times text{width}timestext{height} = 500text{ Pa} times 0.86text{ m}times 1.15text{ m} = 495text{ N}
$$
or in pounds
$$
F = 495text{ N} = 495times 0.225text{ lb} = 111text{ lb}.
$$
b) The total weight of all combined things being supported by buoyancy force remains the same so the same amount of water is being displaced resulting in the water level remaining constant. Even though less water is displaced by the boat when block are thrown out this difference is exactly compensated by the amount of water being displaced by individual floating blocks.
$$
mg=rho V’g
$$
where $V’$ is immersed volume. Knowing that $m=rho_{doughnut}V$ where $V$ is the volume of the doughnut we get
$$
rho_ {doughnut}V g = rho V’g
$$
which yields exactly
$$
frac{V’}{V} =frac{rho_{doughnut}}{rho} = frac{3}{4}.
$$
When we turn the doughnut around the previously upper three quarters will be now immersed. This means that upper and lower three quarters will be fried for the same amount of time but since they obviously overlap the middle half of the doughnut will be fried twice resulting in its’ darker brown color. So the final result is outer quarters will be light brown and the middle half dark brown as explained in C.
$$
mg=rho V’g
$$
where $V’=frac{1-0.22}{2} H B$ is the immersed volume and $B$ is the area of its’ base. Factor of $alpha=(1-0.22)/2$ is the fraction of it’s height being immersed when it is on its’ one side and we got it by solving the equation $0.22+2alpha = 1$ because fractions of heights that are brown ($alpha$) and that of the white stripe ($0.22$) have to add up to one.
Knowing also that $m=rho_{doughnut} V = rho_{doughnut} HB$ where $V$ is doughnuts volume we get
$$
rho_{doughnut}V=rho V’ Rightarrow
$$
which yields
$$
frac{V’}{V} = frac{frac{1-0.22}{2} H}{H} = 0.39 = frac{rho_{doughnut}}{rho}.
$$
Now we easily obtain
$$
rho_{doughnut} = 0.39rho = 360text{ kg/m}^3.
$$
$$
(2alpha+beta)H=H
$$
yielding
$$
beta = 1-2alpha.
$$
Now we only need to determine $alpha$ i.e. the fraction of height submerged when on one side. From the condition of balance between doughnut’s weight and buoyancy we get
$$
rho_{doughnut}Vg = rho V’g.
$$
Knowing that the volume of a cylinder is its’ height multiplied by the area of its’ base and canceling $g$ we get
$$
frac{rho{doughnut}}{rho} = frac{V’}{V} =frac{Balpha H}{BH}=alpha.
$$
Substituting for densities we obtain
$$
alpha = 0.36.
$$
Now putting this into the equation for $beta$ we get
$$
beta = 0.28.
$$
