Examples of V/Q ratio in the following topics:
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- This is referred to as ventilation/perfusion (V/Q) mismatch.
- There are two types of V/Q mismatch that produce dead space.
- This will decrease ventilation but not affect perfusion; therefore, the V/Q ratio changes and gas exchange is affected .
- Pulmonary edema with small pleural effusions on both sides (as shown) can cause changes in the V/Q ratio.
- Compare and contrast anatomical and physiological dead space and their role in V/Q mismatch
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- The ratio of ventilation in the lungs and perfusion of the lungs (the air and blood supply of the alveoli respectively) is called the V/Q ratio (with Q being perfusion); it is an important indicator of efficiency in the lungs.
- The V/Q in a healthy individual is variable but balanced, usually around .8, but can be higher or lower depending on the region of the lung sampled.
- An imbalance in V/Q ratio is called a mismatch, and indicates a severe problem.
- Too low perfusion (and a higher ratio) indicates alveolar dead space, while too low ventilation (and a lower ratio) indicates a shunt, which is a lack of air supply relative to perfusion.
- V/Q mismatches occur during many different kinds of lung failure, and have many different causes.
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- Calculate the total energy radiated per unit time, expressing it in terms of the constants already defined and the ratio $\gamma=1/\sqrt{1-\beta^2}$ of the particle's total energy to its rest energy.
- What is the time if $\displaystyle P = \frac{2 q^2 \dot{u}^2}{3 c^3} = \frac{2 q^2 }{3 c^3} \left ( \frac{q^2}{r^2 m} \right )^2$ A (for an hydrogen)?
- $\displaystyle \dot{\bf u} = -{\hat r} \frac{q^2}{r^2 m}$$P = \frac{2 q^2 \dot{u}^2}{3 c^3} = \frac{2 q^2 }{3 c^3} \left ( \frac{q^2}{r^2 m} \right )^2$
- $\displaystyle E = -\frac{q^2}{r} + \frac{1}{2} m v^2 = -\frac{q^2}{r} + \frac{1}{2} \left ( \frac{q^2}{r} \right ) = -\frac{1}{2} \frac{q^2}{r} $for where I used$\displaystyle \frac{m v^2}{r} = \frac{q^2}{r^2}$circular motion.
- $\displaystyle P = \frac{2 q^2 \dot{u}^2}{3 c^3} = \frac{2 q^2 }{3 c^3} \left ( \frac{q^2}{r^2 m} \right )^2$$\displaystyle P = \frac{2 q^2 \dot{u}^2}{3 c^3} = \frac{2 q^2 }{3 c^3} \left ( \frac{q^2}{r^2 m} \right )^2$
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- $\displaystyle {\bf q} = \left [ \frac{1}{2} V^2 + w \right ] \rho {\bf V} .$
- As we take the limit of a stong shock $(p_2,V_2)$ we find that the compresssion ratio and square of the downstream Mach number approach
- For $\gamma=5/3$ the compression ratio is 4 and the downstream Mach number is $1/\sqrt{5}$.
- For a diatomic gas ($\gamma=7/5$) the maximum compression ratio is larger at 6 and the square of the downstream Mach number is $1/\sqrt{7}$ --- in fact the compression ratio $\rho_2/\rho_1$ always equals $1/M_2^2 - 1$ for any value of$M_1$.
- so the Mach numbers on each side of the shock are given by the ratio of the slope of the secant to the slope of the tangent.
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- Let's expand this ratio for $M_1^2 \approx 1$ to understand the change in entropy for a weak shock,
- where $Q$ is the volumetric flow rate and $r$ is the radius of the jump.
- Now we use that $v_1$ is constant to eliminate $h_1=j/v_1=Q/(2\pi r v_1)$.
- $\displaystyle \frac{Q}{2\pi r v_1} = \frac{Q t}{2 A} \left ( \sqrt{ 1 + \frac{8}{g} \left (\frac{Q}{2\pi r} \right )^2 \left ( \frac{A}{Q t} \right)^3 } - 1 \right ).$
- $\displaystyle r = \frac{A \left ( 2 A v_1^2 - t Q g \right )}{2 \pi t^2 Q g v_1} = \frac{A^2 v_1}{Q g \pi} \frac{1}{t^2} - \frac{A}{2\pi v_1} \frac{1}{t}
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- $\displaystyle \int_V dV \nabla \cdot {\vec E} = \int dV 4 \pi \rho = 4 \pi q$
- $\displaystyle \int_{\partial V} dV \nabla \cdot {\vec E} = \int {\vec E} \cdot d A = |{\vec E}| 4 \pi R^2 = 4 \pi q$
- $\displaystyle F = \frac{m v^2}{r} = \frac{ e v B}{c}, v\approx c, m\approx \frac{E}{c^2}$
- We have $\displaystyle {\vec v} = \frac{Q}{\beta} {\vec E}$.
- The ratio of the momentum unit to the energy unit must be the ratio of the force (momentum per unit time) to the power (energy per unit time), so we get
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- Solution Rearranging I=V/R and substituting known values gives $R=\frac{V}{I}=\frac{12 \ V}{2.5 \ A}=4.8 \ \Omega$.
- Additional insight is gained by solving I=V/R for V, yielding V=IR.
- In a simple circuit (one with a single simple resistor), the voltage supplied by the source equals the voltage drop across the resistor, since E=qΔV, and the same q flows through each.
- That is, the ratio of V/I is constant, and when current is plotted as a function of voltage the curve is linear (a straight line).
- If voltage is forced to some value V, then that voltage V divided by measured current I will equal R.
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- The electric potential of a point charge Q is given by $V=\frac{kQ}{r}$.
- Another way of saying this is that because PE is dependent on q, the q in the above equation will cancel out, so V is not dependent on q.
- Using calculus to find the work needed to move a test charge q from a large distance away to a distance of r from a point charge Q, and noting the connection between work and potential (W=–qΔV), it can be shown that the electric potential V of a point charge is
- Thus V for a point charge decreases with distance, whereas E for a point charge decreases with distance squared:
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- with the relativistic mass m and its charge q.
- Mass spectrometry is an analytical technique that measures the mass-to-charge ratio of charged particles.
- Mass analyzers separate the ions according to their mass-to-charge ratio.
- This differential equation along with initial conditions completely determines the motion of a charged particle in terms of m/Q.
- The deflections of the particles are dependent on the mass-to-charge ratio.
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- An object's heat capacity (symbol C) is defined as the ratio of the amount of heat energy transferred to an object to the resulting increase in temperature of the object .
- A sample containing twice the amount of substance as another sample requires the transfer of twice as much heat (Q) to achieve the same change in temperature (ΔT).
- The values thus measured are usually subscripted (by p and V, respectively) to indicate the definition.