In fluid dynamics, dynamic pressure (denoted by q or Q and sometimes called velocity pressure) is the quantity defined by: [1] = where (in SI units): q is the dynamic pressure in pascals (i.e., N/m 2), ρ (Greek letter rho) is the fluid mass density (e.g. in kg/m 3), and; u is the flow speed in m/s.

In physics and engineering, mass flow rate is the rate at which mass of a substance changes over time. Its unit is kilogram per second (kg/s) in SI units, and slug per second or pound per second in US customary units. The common symbol is (pronounced "m-dot"), although sometimes (Greek lowercase mu) is used.

In quantum mechanics, charge density ρ q is related to wavefunction ψ(r) by the equation = | | where q is the charge of the particle and | ψ(r) | 2 = ψ*(r)ψ(r) is the probability density function i.e. probability per unit volume of a particle located at r.

2016年9月19日 · For example, if we are inside the sphere, we get $ q_\textrm{enc} = \rho V=\frac q {V_1}V_2=\frac {q} {\frac 4 3\pi R^3}(\frac 4 3\pi r^3) =\frac {qr^3} {R^3}$. This will then simplify to the proper electric field, once we set it divided by $\varepsilon_0$ equal to the flux, which is $\vec E =\frac 1 {4\pi \varepsilon_0} \frac {qr} {R^3}$.

2 天之前 · Flow rate Q is defined as the volume V flowing past a point in time t. The SI unit of flow rate is (m^3)/s, but other rates can be used, such as L/min. Flow rate and velocity are related by the …

ρ = q V. Where. The SI unit of Charge density is Coulomb per unit measurement under consideration. Q.1: Determine the charge density of an electric field, if a charge of 6 C per meter is present in a cube of volume 3 m3. Solution: Given parameters are as follows: Electric Charge, q = 6 C per m. Volume of the cube, V = 3 m3.

2020年9月2日 · So far, I have used the 1st law of thermodynamics to state that $\frac{d U}{d t} = \frac{d Q}{d t}$, since the system's volume is constant. Furthermore, from the definition of the heat capacity, I also know that $\Delta Q = \rho V C \Delta T$.

其中 F 可以为 \rho,\theta, E 等物理量。 关于 通量 的计算，以某物理量单位时间流过右侧单位面积面元的通量为例。 得到N-S方程有.

假设流体各处的流速相同且恒为 v ，按照下图所标字母可知流体通过垂直截面的流量为 Q = \frac{{\rho v\Delta t{S_ \bot }}}{{\Delta t}} = \rho v{S_ \bot }\\ 而通过倾斜截面的流量也为 Q = \frac{{\rho v\Delta t{S_ \bot }}}{{\Delta t}} = \rho v{S_ \bot }\\

很多时候，条件给出的都是 电荷体密度 \rho 而非线密度 \lambda ，又该如何替换？ 其实很简单，在脑中想象一个三维空间坐标系 xyz (这里使用右手系)，那么线密度 \lambda 代表的就是沿 z 轴方向的电荷分布，对于题给圆柱而言，要求某段圆柱带电量，只需乘以沿 z 轴方向的长度即可 (前提是要求圆柱的底面半径必须与题给圆柱相同)，即 Q=\lambda h...... (*) ；那么对于体密度 \rho 而言就更好理解，只需用电荷体密度乘以要求圆柱的体积即可： Q=\rho V=\rho \pi R_0^2h......