Answer:
![Q_{cv} = -1007.86kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20-1007.86kJ)
Explanation:
Our values are,
State 1
![V=3m^3\\P_1=1bar\\T_1 = 295K](https://tex.z-dn.net/?f=V%3D3m%5E3%5C%5CP_1%3D1bar%5C%5CT_1%20%3D%20295K)
We know moreover for the tables A-15 that
![u_1 = 210.49kJ/kg\\h_i = 295.17kJkg](https://tex.z-dn.net/?f=u_1%20%3D%20210.49kJ%2Fkg%5C%5Ch_i%20%3D%20295.17kJkg)
State 2
![P_2 =6bar\\T_2 = 296K\\T_f = 320K](https://tex.z-dn.net/?f=P_2%20%3D6bar%5C%5CT_2%20%3D%20296K%5C%5CT_f%20%3D%20320K)
For tables we know at T=320K
![u_2 = 228.42kJ/kg](https://tex.z-dn.net/?f=u_2%20%3D%20228.42kJ%2Fkg)
We need to use the ideal gas equation to estimate the mass, so
![m_1 = \frac{p_1V}{RT_1}](https://tex.z-dn.net/?f=m_1%20%3D%20%5Cfrac%7Bp_1V%7D%7BRT_1%7D)
![m_1 = \frac{1bar*100kPa/1bar(3m^3)}{0.287kJ/kg.K(295k)}](https://tex.z-dn.net/?f=m_1%20%3D%20%5Cfrac%7B1bar%2A100kPa%2F1bar%283m%5E3%29%7D%7B0.287kJ%2Fkg.K%28295k%29%7D)
![m_1 = 3.54kg](https://tex.z-dn.net/?f=m_1%20%3D%203.54kg)
Using now for the final mass:
![m_2 = \frac{p_2V}{RT_2}](https://tex.z-dn.net/?f=m_2%20%3D%20%5Cfrac%7Bp_2V%7D%7BRT_2%7D)
![m_2 = \frac{1bar*100kPa/6bar(3m^3)}{0.287kJ/kg.K(320k)}](https://tex.z-dn.net/?f=m_2%20%3D%20%5Cfrac%7B1bar%2A100kPa%2F6bar%283m%5E3%29%7D%7B0.287kJ%2Fkg.K%28320k%29%7D)
![m_2 = 19.59kg](https://tex.z-dn.net/?f=m_2%20%3D%2019.59kg)
We only need to apply a energy balance equation:
![Q_{cv}+m_ih_i = m_2u_2-m_1u_1](https://tex.z-dn.net/?f=Q_%7Bcv%7D%2Bm_ih_i%20%3D%20m_2u_2-m_1u_1)
![Q_{cv}=m_2u_2-m1_u_1-(m_2-m_1)h_i](https://tex.z-dn.net/?f=Q_%7Bcv%7D%3Dm_2u_2-m1_u_1-%28m_2-m_1%29h_i)
![Q_{cv} = (19.59)(228.42)-(3.54)(210.49)-(19.59-3.54)(295.17)](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20%2819.59%29%28228.42%29-%283.54%29%28210.49%29-%2819.59-3.54%29%28295.17%29)
![Q_{cv} = -1007.86kJ](https://tex.z-dn.net/?f=Q_%7Bcv%7D%20%3D%20-1007.86kJ)
The negative value indidicates heat ransfer from the system
Answer: Last week, Nate and I counted all the inventory.
Explanation: all other choices are passive voices
this sentence follows a clear subject + verb + object construct that's why it is an active voice. In fact, sentences constructed in the active voice add impact to your writing. but on the other hand With passive voice, the subject is acted upon by the verb.
Ape x
Explanation:
We need to rearrange the following formula for the values given in parenthesis.
(1) x+xy = y, (x)
taking x common in LHS,
x(1+y)=y
![x=\dfrac{y}{1+y}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7By%7D%7B1%2By%7D)
(2) x+y = xy, (x)
Subrtacting both sides by xy.
x+y-xy = xy-xy
x+y-xy = 0
x-xy=-y
x(1-y)=-y
![x=\dfrac{-y}{1-y}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7B-y%7D%7B1-y%7D)
(3) x = y+xy, (x)
Subrating both sides by xy
x-xy = y+xy-xy
x(1-y)=y
![x=\dfrac{y}{1-y}](https://tex.z-dn.net/?f=x%3D%5Cdfrac%7By%7D%7B1-y%7D)
(4) E = (1/2)mv^2-(1/2)mu^2, (u)
Subtracting both sides by (1/2)mv^2
E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2
E-(1/2)mv^2 =-(1/2)mu^2
So,
![2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}](https://tex.z-dn.net/?f=2%28E-%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%29%3D-mu%5E2%5C%5C%5C%5Cu%5E2%3D%5Cdfrac%7B-2%7D%7Bm%7D%28E-%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%29%5C%5C%5C%5Cu%3D%5Csqrt%7B%5Cdfrac%7B-2%7D%7Bm%7D%28E-%5Cdfrac%7B1%7D%7B2%7Dmv%5E2%29%7D%5C%5C%5C%5Cu%3D%5Csqrt%7B%5Cdfrac%7B2%7D%7Bm%7D%28%5Cdfrac%7B1%7D%7B2%7Dmv%5E2-E%29%7D)
(5) (x^2/a^2)-(y^2/b^2) = 1, (y)
![\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D-1%3D%5Cdfrac%7By%5E2%7D%7Bb%5E2%7D%5C%5C%5C%5Cy%5E2%3Db%5E2%28%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D-1%29%5C%5C%5C%5Cy%3Db%5Csqrt%7B%5Cdfrac%7Bx%5E2%7D%7Ba%5E2%7D-1%7D)
(6) ay^2 = x^3, (y)
![y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}](https://tex.z-dn.net/?f=y%5E2%3D%5Cdfrac%7Bx%5E3%7D%7Ba%7D%5C%5C%5C%5Cy%3D%5Csqrt%7B%5Cdfrac%7Bx%5E3%7D%7Ba%7D%7D)
Hence, this is the required solution.
Answer:
ay max = 4.91 m/s²
so here acceleration would be either right or left
Explanation:
given data
wide b = 1 m
long l = 2 m
depth d = 3 m
height of tank sides h = 4 m
solution
here for prevent spilling condition is
≤
..........1
≤ - 0.50
and when here
= -
......2
when az is 0 ay will be
ay = -
and ay max will be
ay max = -( -0.50) (9.81 )
ay max = 4.91 m/s²
so here acceleration would be either right or left