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The exit speed is determined by the hydrostatic pressure at the hole, which depends on the vertical height of water above the hole. Less water above the hole → lower pressure → lower exit speed.
To mathematically prove the solution, we use Bernoulli’s equation to understand how the speed of the water as it exits the hole changes with the distance \( y \).
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}\) | Bernoulli’s equation relates pressure \( P \), fluid velocity \( v \), and height \( h \), assuming steady flow of an incompressible fluid with no friction. |
| 2 | \(P_{\text{surface}} = P_{\text{hole}} + \frac{1}{2} \rho v^2 + \rho g(d-y)\) | Apply Bernoulli’s equation from the water surface (height \( d \)) to the hole (height \( y \)). |
| 3 | \(P_{\text{surface}} = P_{\text{hole}}\) | The pressures at the surface and the hole are both atmospheric pressure, so they cancel each other out. |
| 4 | \(\frac{1}{2} \rho v^2 = \rho g(d-y)\) | Rearrange the equation after canceling out pressure terms. This relates the kinetic energy per unit volume to the gravitational potential energy per unit volume. |
| 5 | \(v = \sqrt{2g(d-y)}\) | Solve for the velocity \( v \) of water exiting the hole. |
| 6 | \(\text{As } y \text{ increases, } d-y \text{ decreases} \Rightarrow v \text{ decreases}\) | The velocity of water decreases as the distance \( y \) increases because the effective height of the water column (\( d-y \)) decreases, reducing the pressure difference driving the flow. |
The correct answer is:
(a) The speed would decrease because the water pressure at the depth of the hole would decrease.
Explanation for Incorrect Options:
– (b) The air pressure just outside of the hole is not a factor in the velocity, as it remains constant.
– (c) More gravitational potential energy is not being converted to kinetic energy as \( y \) increases because the height of the water column contributing to the speed is decreasing.
– (d) There is no additional upward push from the water below affecting the horizontal speed at the hole; the speed depends on the depth above the hole.
Just ask: "Help me solve this problem."
A small rock sits at the bottom of a cup filled with water. The upward force exerted by the water on the rock is \( F_0 \). The water is then poured out and replaced by an oil that is \( \frac{3}{4} \) as dense as water, and the rock again sits at the bottom of the cup, completely under the oil. Which of the following expressions correctly represents the magnitude of the upward force exerted by the oil on the rock?
Two blocks of the same size are floating in a container of water. The first block is submerged \( 80\% \) while the second block is submerged by \( 20\% \) beneath the water. Which of the following is a correct statement about the two blocks?
A fountain with an opening of radius \( 0.015 \) \( \text{m} \) shoots a stream of water vertically from ground level at \( 6.0 \) \( \text{m/s} \). The density of water is \( 1000 \) \( \text{kg/m}^3 \).
The figure shows a horizontal pipe with sections with different cross-sectional areas. Small tubes extend from the top of each section. The cross-sectional area of the pipe at location C is half that at A, and the areas at A and D are the same. Water flows in the pipe from left to right. Which of the following correctly ranks the height \( h \) of the water in the tubes above the labeled locations?
You have a giant cask of water with a spigot some height below the water surface. The surface of the water, which is essentially at rest, is exposed to atmosphere (\( \approx 10^5 \text{Pa} \)). The water density is \( \approx 1000 \text{kg/m}^3 \). The water pours out of the spigot at \( 3 \text{m/s} \). How far below the water surface is the spigot positioned?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.Â
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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