Examples of equation in the following topics:
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- There are four kinematic equations that describe the motion of objects without consideration of its causes.
- Notice that the four kinematic equations involve five kinematic variables: $d$, $v$, $v_0$, $a$, and $t$.
- Each of these equations contains only four of the five variables and has a different one missing.
- Step two - Find an equation or set of equations that can help you solve the problem.
- Choose which kinematics equation to use in problems in which the initial starting position is equal to zero
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- We can prove this simply by integrating the fourth equation over $d^3 {\bf p}$.
- The first two terms yield the left-hand side of the equation above.
- Let's define ${\bf V}=\langle {\bf v} \rangle$ and write out the equation above by components,
- This is the continuity equation.
- Because the fourth equation is consistent with the Lioville equation (seventh equation) and more generally with the Boltzmann equation (sixth equation) and $J^\mu_{;\mu}=0$ if particles are conserved, the Lioville and Boltzmann equations cannot hold if $\nabla_{\bf p} \cdot F \neq 0$ and particles are conserved.
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- Identify the problem and solve the appropriate equation or equations for the quantity to be determined.
- Solve the appropriate equation or equations for the quantity to be determined (the unknown).
- We cannot use any equation that incorporates t to find ω, because the equation would have at least two unknown values.
- The equation $\omega 2=\omega 02+2$ will work, because we know the values for all variables except ω.
- Taking the square root of this equation and entering the known values gives
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- The wavefunction evolves forward in time according to the time-dependent Schrodinger equation
- If the Hamiltonian is independent of time we can solve this equation by
- This realization allows us to write the equation that the wavefunction of an atom must satisfy
- For most atomic states, these effects can be treated at perturbations.We can simplify these equations by using
- This gives the following dimensionless equation
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- The Ideal Gas Law is the equation of state of a hypothetical ideal gas.
- Variations of the ideal gas equation may help solving the problem easily.
- Substitute the known values into the equation.
- Choose a relevant gas law equation that will allow you to calculate the unknown variable: We can use the general gas equation to solve this problem: $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$.
- Substitute the known values into the equation.
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- A next step is to use what is known to find the appropriate equation to find what is unknown.
- While it is easiest to find an equation that leaves only one unknown, sometimes this is not possible.
- In these situations, you can solve multiple equations to find the right answer.
- Remember that equations represent physical principles and relationships, so use the equations and drawings in tandem.
- You may then substitute the knowns into the appropriate equations and find a numerical solution.
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- We can write $\displaystyle P' = (\partial P/\partial \rho)_s \rho'$ and rewrite the continuity equation to get
- This is a wave equation with a sound speed of $c_s^2 = (\partial P/\partial \rho)_s$.
- Let's take a solution to this equation for the pressure,
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- This simply adds another term to the above version of the Bernoulli equation and results in
- Bernoulli's equation can be applied when syphoning fluid between two reservoirs .
- The Bernoulli equation can be adapted to flows that are both unsteady and compressible.
- However, the assumption of inviscid flow remains in both the unsteady and compressible versions of the equation.
- Adapt Bernoulli's equation for flows that are either unsteady or compressible
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- is a solution of the wave equation for $c= \frac{\omega}{k}$.
- is a solution to the wave equation.
- To solve the equation 1, let's introduce new variables: $\phi = x-ct, \psi= x+ct$.
- With the change of variables, the equation 1 becomes $\frac{\partial u_+}{\partial \phi} = 0$ for the equation with the "+" sign and $\frac{\partial u_-}{\partial \psi} = 0$ for the "-" sign.
- Any function that contains "x+ct" or "x-ct" can be a solution of the wave equation.
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- So the vector differential equation governing this simple harmonic motion is:
- In other words this one vector equation is equivalent to three completely separate scalar equations (using $\omega _0 ^2 = k/m$ )
- The equations are uncoupled in the sense that each unknown ( $x,y,z$ ) occurs in only one equation; thus we can solve for $x$ ignoring $y$ and $z$ .
- Plugging these into the equations for $x$ and $y$ gives the two amplitude equations
- We can use the first equation to compute $x_0$ in terms of $y_0$ and then plug this into the second equation to get