Question
In the following diagram if $\mathrm{V}_{2}>\mathrm{V}_{1}$ then....(A) $\lambda_{1}=\sqrt{\lambda_{2}}$(B) $\lambda_{1}<\lambda_{2}$(C) $\lambda_{1}=\lambda_{2}$(D) $\lambda_{1}>\lambda_{2}$
Step 1
7 \times 10^{-26}kg$, and $P=18Pa$. We also know that the energy $E$ is given by $E=\frac{1}{2}mv^{2}=3kT$. Show more…
Show all steps
Your feedback will help us improve your experience
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The values of $\lambda$ for which the equations $3 \mathrm{x}^{2}-2 \lambda \mathrm{x}-4=0$ and $\mathrm{x}^{2}-4 \lambda \mathrm{x}+2=0$ have a common root are (a) $\frac{1}{2}, \frac{-1}{2}$ (b) $\frac{1}{\sqrt{2}}, \frac{-1}{\sqrt{2}}$ (c) $\frac{1}{\sqrt{5}}, \frac{-1}{\sqrt{5}}$ (d) $\frac{1}{4},-\frac{1}{4}$
If $\vec{B}=\lambda \vec{A}$ then $\frac{\vec{B}}{\vec{A}}=\ldots$ (A) $\lambda$ (B) $\frac{1}{\lambda}$ (C) $\frac{\lambda}{2}$ (D) Indeterminate
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD