All Solutions
Page 451: Standardized Test Prep
$$
begin{gather*}
m , g = F_{text{buoyant}} \
rho_{text{ball}} cdot V cdot g = rho_{text{water}} cdot frac{V}{2} cdot g
end{gather*}
$$
From which we deduce:
$$
begin{align*}
rho_{text{ball}} = frac{rho_{text{water}} }{2}
end{align*}
$$
From this we see that the statement A is incorrect.
Obviously statements B and C are incorrect since the forces acting on the ball are balanced.
We conclude that statement D is correct.
Secondly conclude that the gas pressure is always equal to the static pressure in the water surrounding it for that given depth. Knowing this we can write:
$$
begin{align*}
P(h_1) cdot V_1 = P(h_2) cdot V_2
end{align*}
$$
Our two heights are $h_1 = 21text{ m}$ and $h_2 = 0text{ m}$.
Now we find the pressures by increasing the atmospheric pressure for the amount of hydrostatic pressure as follows:
$$
begin{align*}
P(h) = P_{text{atm}} + rho cdot g cdot h
end{align*}
$$
Inserting the numbers we have:
$$
begin{align*}
P(h_1) &= 101.3text{ kPa} + 1000 ; frac{text{kg}}{text{m}^3} cdot 9.81 ; frac{text{m}}{text{s}^2} cdot 21text{ m} = 307.31text{ kPa} \
P(h_2) &= 101.3text{ kPa}
end{align*}
$$
Now we find the volume $V_2$:
$$
begin{align*}
V_2 &= V_1 cdot frac{P(h_1)}{P(h_2)} \
V_2 &= 0.001text{ m}^3 cdot frac{307.31text{ kPa} }{101.3text{ kPa}} \
V_2 &approx 0.003text{ m}^3
end{align*}
$$
It states:
$$
begin{equation*}
A cdot v = text{const}
end{equation*}
$$
Secondly, we know that the ideal gas equation reads:
$$
begin{equation*}
P cdot V = N cdot k cdot T
end{equation*}
$$
Which means that the pressure is proportional to $T$ when the volume and number of particles are fixed, as they are in our sealed, rigid container.
So it follows that B and C are incorrect.
D is obviously incorrect since $Psim T$
The correct answer is A.
This could also have been concluded from the Equation of Continuity.
We know that the static pressure is lower in regions where the speed of the fluid is greater, this stems from Bernoulli’s equation. Notice that the pipe was horizontal, so we didn’t have to take into consideration the gravitational potential term in the equation.
We conclude that the static pressure decreases in the narrow part of the pipe since the speed of the fluid flow is greater there.
– The volume of the container $V = 0.022text{ m}^3$
– The pressure of the container $P = 310 text{ kPa}$
– The temperature of the container $T = 25text{textdegree}text{C} = 298.15text{ K}$
To calculate the number of helium atoms we use the Ideal Gas Equation:
$$
begin{equation*}
PV = N , k , T
end{equation*}
$$
rearranging for number of particles:
$$
begin{equation*}
N = frac{PV}{k , T}
end{equation*}
$$
Plugging in the values we get:
$$
begin{align*}
N = frac{310 cdot 10^{3}text{ Pa} cdot 0.022text{ m}^3 }{1.38 cdot 10^{-23} ; frac{text{J}}{text{K}} cdot 298.15text{ K} } = 1.66 cdot 10^{24}
end{align*}
$$
The number of helium atoms in the balloon is $N = 1.66 cdot 10^{24}$, which is approximately $1.7 cdot 10^{24}$.
begin{enumerate}[A)]
item
$N = 1.7 cdot 10^{24}$
end{enumerate}
$$
begin{gather*}
m , g = F_{text{buoyant}} \
rho_{text{object}} cdot V cdot g = rho_{text{water}} cdot 0.75 V cdot g
end{gather*}
$$
From which we deduce:
$$
begin{align*}
rho_{text{object}} = 0.75 , rho_{text{water}}
end{align*}
$$
Now renaming $rho_{text{water}} = rho$
We see that statement B is the correct one.
First remember, the spring constant is defined as:
$$
begin{equation*}
k = frac{F}{x}
end{equation*}
$$
Where $F$ is the force pulling on the spring, and $x$ is its elongation.
Now consider the following, if we joined together end to end two springs of half length, such as the new one we get by cutting a spring in half, we would construct the initial spring.
When one pulls on such a system of two joined springs he would get that the same force $F$ stretches both springs, and that each of them is extended. Call these elongations $x_1$ and $x_2$, but be aware that $x_1 = x_2$ since the spring are the same.
So the effective coefficient of the system would be:
$$
begin{align*}
k_{_{eff}} = frac{F}{(x_1 + x_2)}
end{align*}
$$
Where you could also write the individual equations:
$$
begin{align*}
k_1 &= frac{F}{x_1} \
k_2 &= frac{F}{x_2}
end{align*}
$$
In our case the situation gets simplified drastically by $k_1 = k_2$ and $x_1 = x_2$ since the springs are the same, so we have:
$$
begin{align*}
k_{text{eff}} = frac{F}{2x_1} = frac{1}{2} cdot k_1
end{align*}
$$
We conclude that the coefficient of stiffness of the total spring is half as big as the one associated with the half spring.
That is, by cutting a spring in half, we increase the stiffness constant of each new half length spring by two
The correct answer is C) $2k$
$$
begin{equation*}
P = P_{text{atm}} + frac{m , g}{A}
end{equation*}
$$
Where $m$ is the mass of the piston, and $A$ its area.
The initial pressure in the container $P_{text{i}}$ is found by using $m = 2text{ kg}$.
When another mass is added the total mass becomes $m = 2text{ kg} + 1text{ kg} = 3text{ kg}$.
Obviously this will result in a larger pressure $P_{text{f}}$:
$$
begin{align*}
P_{text{f}} = P_{text{i}} + frac{1text{ kg}}{A}
end{align*}
$$
$$
begin{equation*}
P cdot V = text{const}
end{equation*}
$$
Now applying it for the initial and final positions we get:
$$
begin{align*}
P_{text{i}} cdot V_{text{i}} = P_{text{f}} cdot V_{text{f}}
end{align*}
$$
Rearranging:
$$
begin{align*}
V_{text{f}} = V_{text{i}} cdot frac{P_{text{i}} }{P_{text{f}} }
end{align*}
$$
Since we know that the pressure increases, we see that $frac{P_{text{i}} }{P_{text{f}}} < 1$ so $V_{text{f}} < V_{text{i}}$
This could have been reasoned by common sense as well.
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