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题目

解题步骤

1. Understand the question

   Turn 1 Question

   assessment

   1. use dimensional analysis to show that the equation ( v = (2gh)^{\frac{1}{2}} ) is dimensionally correct where

   ( v
   ightarrow ) velocity, ( g
   ightarrow ) acceleration due to gravity, ( h
   ightarrow ) height.

   1. velocity of deep water waves relates to the density of the water, the acceleration due to gravity, ( g ), and the wave length, ( lambda ). establish an equation using dimensional analysis.
   2. the real gas equation is given by ( left( p + \frac{a}{v^2}

   ight)(v - b) = rt ), where ( p ) is pressure, ( v ) is the volume, ( t ) is the temperature, ( r ) is the gas constant, ( a ) and ( b ) are constants. what are the dimensions of ( a ) and ( b ).

   1. given ( f = \frac{k m_1 m_2}{r^2} ), ( f = ) force, ( m_1 ) and ( m_2 ) are masses, ( r ) is the distance between ( m_1 ) and ( m_2 ). find the dimensions and unit of ( k ).

   Turn 2 Question

   Explain q1

   Turn 3 Question

   Explain 2

2. Response

   Turn 1 Answer

3. Explanation

   Step1: Define base dimensions

   Let $[M]$ = mass, $[L]$ = length, $[T]$ = time.

   ---

   For Question 1:

   Step1: Write dimensions of each quantity

   $[v] = LT^{-1}$, $[g] = LT^{-2}$, $[h] = L$

   Step2: Compute RHS dimensions

   $$[(2gh)^\frac{1}{2}] = ([LT^{-2}] \cdot [L])^\frac{1}{2} = (L^2T^{-2})^\frac{1}{2} = LT^{-1}$$

   Step3: Compare LHS and RHS

   $[v] = [(2gh)^\frac{1}{2}] = LT^{-1}$

   ---

   For Question 2:

   Step1: Assume proportionality relation

   Let $v = k
   ho^x g^y \lambda^z$, $k$ = dimensionless constant.

   Step2: Substitute dimensions

   $$LT^{-1} = [ML^{-3}]^x [LT^{-2}]^y [L]^z = M^x L^{-3x+y+z} T^{-2y}$$

   Step3: Equate exponents

   • Mass: $x=0$
   • Time: $-2y=-1 \implies y=\frac{1}{2}$
   • Length: $-3(0)+\frac{1}{2}+z=1 \implies z=\frac{1}{2}$

   Step4: Write final equation

   $v = k\sqrt{g\lambda}$

   ---

   For Question 3:

   Step1: Analyze dimension of $b$

   $b$ adds to $V$, so $[b] = [V] = L^3$

   Step2: Analyze dimension of $a$

   $\frac{a}{V^2}$ has same dimension as $P$: $[P] = ML^{-1}T^{-2}$
   $$[a] = [P][V^2] = ML^{-1}T^{-2} \cdot (L^3)^2 = ML^5T^{-2}$$

   ---

   For Question 4:

   Step1: Rearrange for $k$

   $k = \frac{F r^2}{m_1 m_2}$

   Step2: Substitute dimensions

   $$[k] = \frac{[MLT^{-2}] \cdot [L^2]}{[M][M]} = M^{-1}L^3T^{-2}$$

   Step3: Find SI unit

   Unit: $\frac{\text{N} \cdot \text{m}^2}{\text{kg}^2} = \text{m}^3 \text{kg}^{-1} \text{s}^{-2}$

4. Final answer
   1. The equation is dimensionally correct, as both sides have dimensions $LT^{-1}$.
   2. The velocity equation is $v = k\sqrt{g\lambda}$ (where $k$ is a dimensionless constant).
   3. Dimensions of $a$: $ML^5T^{-2}$; Dimensions of $b$: $L^3$.
   4. Dimensions of $k$: $M^{-1}L^3T^{-2}$; SI unit of $k$: $\text{m}^3 \text{kg}^{-1} \text{s}^{-2}$ (or $\text{N} \cdot \text{m}^2 \text{kg}^{-2}$).

   Turn 2 Answer

   Oh, for sure! Dimensional analysis is basically just a fancy way of making sure the "units" on both sides of an equals sign actually match up. If they don't, the physics is broken.

   For that first one, you're looking at $v = \sqrt{2gh}$. Here’s how you break it down:

   1. The Left Side ($v$): Velocity is just distance over time, so its dimensions are $L/T$ (Length divided by Time).
   2. The Right Side ($\sqrt{2gh}$):

   • The $2$ is just a number, so we ignore it.
   • $g$ (acceleration) is $L/T^2$.
   • $h$ (height) is just $L$.
   • When you multiply them ($g \times h$), you get $(L/T^2) \times L$, which is $L^2/T^2$.

   1. The Square Root: Since the whole thing is under a square root, you take the root of $L^2/T^2$, which brings you right back to $L/T$.

   Since both sides end up being $L/T$, the equation is "dimensionally correct." It’s like checking if you're comparing apples to apples!

   Does that make the "L" and "T" stuff feel a bit more chill, or is it still looking like alphabet soup?

   Turn 3 Answer

   This one is like a little puzzle where you have to figure out the "recipe" for wave speed. You basically assume the velocity ($v$) depends on density ($
   ho$), gravity ($g$), and wavelength ($\lambda$), and then you use the units to find the exponents.

   Here’s the breakdown:

   1. The Goal: We want $v =

   ho^x \cdot g^y \cdot \lambda^z$. We just need to find $x, y,$ and $z$.

   1. The "Ingredients":

   • Velocity ($v$) is $L/T$.
   • Density ($

   ho$) is $M/L^3$ (Mass/Volume).

   • Gravity ($g$) is $L/T^2$.
   • Wavelength ($\lambda$) is $L$.

   1. Matching them up:

   • Mass ($M$): Velocity has no mass in it, but density does. The only way to get rid of the mass is if $x = 0$. So, density actually doesn't affect the speed!
   • Time ($T$): Velocity has $T^{-1}$ and gravity has $T^{-2}$. To turn a $-2$ into a $-1$, you have to take the square root. So, $y = 1/2$.
   • **Length…

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