The mass of the sour cream that should be added is 0.68 kg.
<h3>Mass of the water</h3>
Mass of the water is calculated as follows;
mass = density x volume
- density of water = 1 kg/L
mass = (1 kg/L) x 0.5 L = 0.5 kg
<h3>Conservation of energy</h3>
Heat gained by the sour cream = Heat lost by the water
![m_s c_s\Delta \theta _s = m_wc_w\Delta \theta _w\\\\m_s = \frac{m_wc_w\Delta \theta _w}{c_s\Delta \theta _s} \\\\](https://tex.z-dn.net/?f=m_s%20c_s%5CDelta%20%5Ctheta%20_s%20%3D%20m_wc_w%5CDelta%20%5Ctheta%20_w%5C%5C%5C%5Cm_s%20%3D%20%5Cfrac%7Bm_wc_w%5CDelta%20%5Ctheta%20_w%7D%7Bc_s%5CDelta%20%5Ctheta%20_s%7D%20%5C%5C%5C%5C)
where;
- Cs is specific heat of sour cream
- Cw is specific heat of water
- Δθ is change in temperature
![m_s = \frac{0.5 \times 4184 \times (80 - 40)}{3510 \times (40 - 5)} \\\\m_s = 0.68 \ kg](https://tex.z-dn.net/?f=m_s%20%3D%20%5Cfrac%7B0.5%20%5Ctimes%204184%20%5Ctimes%20%2880%20-%2040%29%7D%7B3510%20%5Ctimes%20%2840%20-%205%29%7D%20%5C%5C%5C%5Cm_s%20%3D%200.68%20%5C%20kg)
Learn more about heat capacity here: brainly.com/question/16559442
Answer:
Speed, v = 6.91 m/s
Explanation:
Given that,
Spring constant, k = 594 N/m
It is attached to a table and is compressed down by 0.196 m, x = 0.196 m
We need to find the speed of the spring when it is released. Here, the elastic potential energy is balanced by the kinetic energy of the spring such that,
![\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dkx%5E2%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![v=\sqrt{\dfrac{kx^2}{m}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7Bkx%5E2%7D%7Bm%7D%7D)
![v=\sqrt{\dfrac{594\ N/m\times (0.196\ m)^2}{0.477\ kg}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%5Cdfrac%7B594%5C%20N%2Fm%5Ctimes%20%280.196%5C%20m%29%5E2%7D%7B0.477%5C%20kg%7D%7D)
v = 6.91 m/s
So, the speed of the ball is 6.91 m/s. Hence, this is the required solution.
A billiard ball moves with 3 kg⋅m/s of momentum and strikes three other billiard balls that have been just sitting there at rest and not moving.
The total momentum of all four balls after the collision is <em>3 kg⋅m/s</em>, because momentum is not created or destroyed. The total amount of it after an event is the same as the total amount of it before the event.
a. I've attached a plot of the surface. Each face is parameterized by
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b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.
![\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k](https://tex.z-dn.net/?f=%5Cmathbf%20n_1%3D%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_1%7D%7B%5Cpartial%20y%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_1%7D%7B%5Cpartial%20x%7D%3D-%5Cmathbf%20k)
![\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j](https://tex.z-dn.net/?f=%5Cmathbf%20n_2%3D%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_2%7D%7B%5Cpartial%20u%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_2%7D%7B%5Cpartial%20v%7D%3D-u%5C%2C%5Cmathbf%20j)
![\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i](https://tex.z-dn.net/?f=%5Cmathbf%20n_3%3D%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_3%7D%7B%5Cpartial%20z%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_3%7D%7B%5Cpartial%20y%7D%3D-%5Cmathbf%20i)
![\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j](https://tex.z-dn.net/?f=%5Cmathbf%20n_4%3D%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_4%7D%7B%5Cpartial%20v%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_4%7D%7B%5Cpartial%20u%7D%3Du%5C%2C%5Cmathbf%20i%2Bu%5C%2C%5Cmathbf%20j)
![\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k](https://tex.z-dn.net/?f=%5Cmathbf%20n_5%3D%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_5%7D%7B%5Cpartial%20y%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cmathbf%20s_5%7D%7B%5Cpartial%20u%7D%3D2%5Ccos%20u%5C%2C%5Cmathbf%20i%2B2%5Csin%20u%5C%2C%5Cmathbf%20k)
Then integrate the dot product of <em>f</em> with each normal vector over the corresponding face.
![\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_1%7D%5Cmathbf%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Cint_0%5E2%5Cint_0%5E%7B6-x%7Df%28x%2Cy%2C0%29%5Ccdot%5Cmathbf%20n_1%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx)
![=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Cint_0%5E%7B6-x%7D0%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx%3D0)
![\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_2%7D%5Cmathbf%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Cint_0%5E2%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%5Cmathbf%20f%28u%5Ccos%20v%2C0%2Cu%5Csin%20v%29%5Ccdot%5Cmathbf%20n_2%5C%2C%5Cmathrm%20dv%5C%2C%5Cmathrm%20du)
![\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8](https://tex.z-dn.net/?f=%5Cdisplaystyle%3D%5Cint_0%5E2%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D-u%5E2%282%5Csin%20v%2B%5Ccos%20v%29%5C%2C%5Cmathrm%20dv%5C%2C%5Cmathrm%20du%3D-8)
![\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_3%7D%5Cmathbf%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Cint_0%5E2%5Cint_0%5E6%5Cmathbf%20f%280%2Cy%2Cz%29%5Ccdot%5Cmathbf%20n_3%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dz)
![=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Cint_0%5E60%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dz%3D0)
![\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_4%7D%5Cmathbf%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Cint_0%5E2%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%5Cmathbf%20f%28u%5Ccos%20v%2C6-u%5Ccos%20v%2Cu%5Csin%20v%29%5Ccdot%5Cmathbf%20n_4%5C%2C%5Cmathrm%20dv%5C%2C%5Cmathrm%20du)
![=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E2%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D-u%5E2%282%5Csin%20v%2B%5Ccos%20v%29%5C%2C%5Cmathrm%20dv%5C%2C%5Cmathrm%20du%3D%5Cfrac%7B40%7D3%2B6%5Cpi)
![\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_5%7D%5Cmathbf%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%5Cint_0%5E%7B6-2%5Ccos%20u%7D%5Cmathbf%20f%282%5Ccos%20u%2Cy%2C2%5Csin%20u%29%5Ccdot%5Cmathbf%20n_5%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20du)
![=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%5Cint_0%5E%7B6-2%5Ccos%20u%7D12%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20du%3D36%5Cpi-24)
c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.
Alternatively, since <em>S</em> is closed, we can find the total flux by applying the divergence theorem.
![\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_S%5Cmathbf%20f%28x%2Cy%2Cz%29%5Ccdot%5Cmathrm%20d%5Cmathbf%20S%3D%5Ciiint_R%5Cmathrm%7Bdiv%7D%5Cmathbf%20f%28x%2Cy%2Cz%29%5C%2C%5Cmathrm%20dV)
where <em>R</em> is the interior of <em>S</em>. We have
![\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7](https://tex.z-dn.net/?f=%5Cmathrm%7Bdiv%7D%5Cmathbf%20f%28x%2Cy%2Cz%29%3D%5Cdfrac%7B%5Cpartial%283x%29%7D%7B%5Cpartial%20x%7D%2B%5Cdfrac%7B%5Cpartial%28x%2By%2B2z%29%7D%7B%5Cpartial%20y%7D%2B%5Cdfrac%7B%5Cpartial%283z%29%7D%7B%5Cpartial%20z%7D%3D7)
The integral is easily computed in cylindrical coordinates:
![\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%28r%2Ct%29%3Dr%5Ccos%20t%5C%5Cy%28r%2Ct%29%3D6-r%5Ccos%20t%5C%5Cz%28r%2Ct%29%3Dr%5Csin%20t%5Cend%7Bcases%7D%2C0%5Cle%20r%5Cle%202%2C0%5Cle%20t%5Cle%5Cdfrac%5Cpi2)
![\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E2%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%5Cint_0%5E%7B6-r%5Ccos%20t%7D7r%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dt%5C%2C%5Cmathrm%20dr%3D42%5Cpi-%5Cfrac%7B56%7D3)
as expected.
In plant cells, the vacuoles are much larger than in animal cells. When a plant cell has stopped growing, there is usually one very large vacuole. Sometimes that vacuole can take up more than half of the cell's volume.