Answer:
a) the object floats
b) the object floats
c) the object sinks
Explanation:
when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water
a)
volumen for a cube
V=L^3
L=1.53in=0.0388m
V=0.0388 ^3=5.8691x10^-5m^3=58.69ml
density=m/v
density=13.5g/58.69ml=0.23 g/ml
The wooden block floats because it is less dense than water
b)
m=111mg=0.111g
density=m/v
density=0.111g/0.296ml=0.375g/ml
the metal paperclip floats because it is less dense than water
c)
V=0.93cups=220.0271ml
m=0.88lb=399.1613g
Density=m/v
density=399.1613/220.027ml=1.8141g/ml
the apple sinks because it is denser than water
Answer:
please give brainliest my brother just got the corona virus
Explanation:
this is my brothers account he wants to get 5 brainliest
Answer:
The process is possible:
Explanation:
We are going to find out if the entropy generated is greater than 0, if it is greater than 0, then the process is feasible. If it is not, the process is not feasible.
![P_{1} = 3 MPa](https://tex.z-dn.net/?f=P_%7B1%7D%20%3D%203%20MPa)
![x_{1} = 50 % = 0.5](https://tex.z-dn.net/?f=x_%7B1%7D%20%3D%2050%20%25%20%3D%200.5)
![V_{1} = 20 L = 0.02 m^{3}](https://tex.z-dn.net/?f=V_%7B1%7D%20%3D%2020%20L%20%3D%200.02%20m%5E%7B3%7D)
![P_{2} = 1.2 MPa](https://tex.z-dn.net/?f=P_%7B2%7D%20%3D%201.2%20MPa)
![T_{H} = 300^{0} C = 573 K](https://tex.z-dn.net/?f=T_%7BH%7D%20%3D%20300%5E%7B0%7D%20C%20%3D%20573%20K)
Received heat energy, ![Q_{12} = 600 kJ](https://tex.z-dn.net/?f=Q_%7B12%7D%20%3D%20600%20kJ)
Work done, ![W_{12} = 124 kJ](https://tex.z-dn.net/?f=W_%7B12%7D%20%3D%20124%20kJ)
At state 1, using the steam table:
![T_{1} = T_{s} = 233.9^{0} C\\v_{f1} = 0.001216 m^{3} /kg\\v_{fg1} = 0.06546m^{3} /kg\\u_{f1} = 1004.76 kJ/kg\\u_{fg1} = 1599.34 kJ/kg\\s_{f1} = 2.6456 kJ/kg-K\\s_{fg1} = 3.5412kJ/kg-K](https://tex.z-dn.net/?f=T_%7B1%7D%20%3D%20T_%7Bs%7D%20%3D%20233.9%5E%7B0%7D%20C%5C%5Cv_%7Bf1%7D%20%3D%200.001216%20m%5E%7B3%7D%20%2Fkg%5C%5Cv_%7Bfg1%7D%20%3D%200.06546m%5E%7B3%7D%20%2Fkg%5C%5Cu_%7Bf1%7D%20%3D%201004.76%20kJ%2Fkg%5C%5Cu_%7Bfg1%7D%20%3D%201599.34%20kJ%2Fkg%5C%5Cs_%7Bf1%7D%20%3D%202.6456%20kJ%2Fkg-K%5C%5Cs_%7Bfg1%7D%20%3D%203.5412kJ%2Fkg-K)
![v_{1} = v_{f1} + x_{1} * v_{fg1}](https://tex.z-dn.net/?f=v_%7B1%7D%20%3D%20v_%7Bf1%7D%20%2B%20x_%7B1%7D%20%2A%20v_%7Bfg1%7D)
![v_{1} = 0.001216 + 0.5*(0.06546)\\v_{1} = 0.03395 m^{3} /kg](https://tex.z-dn.net/?f=v_%7B1%7D%20%3D%200.001216%20%2B%200.5%2A%280.06546%29%5C%5Cv_%7B1%7D%20%3D%200.03395%20m%5E%7B3%7D%20%2Fkg)
![M = \frac{V_{1} }{v_{1} } \\M = 0.02/0.03395\\M = 0.5892 kg](https://tex.z-dn.net/?f=M%20%3D%20%5Cfrac%7BV_%7B1%7D%20%7D%7Bv_%7B1%7D%20%7D%20%5C%5CM%20%3D%200.02%2F0.03395%5C%5CM%20%3D%200.5892%20kg)
![u_{1} = u_{f1} + x_{1} * u_{fg1}\\u_{1} = 1004.76 + 0.5*1599.34\\u_{1} = 1804.43 kJ/kg](https://tex.z-dn.net/?f=u_%7B1%7D%20%3D%20u_%7Bf1%7D%20%2B%20x_%7B1%7D%20%2A%20u_%7Bfg1%7D%5C%5Cu_%7B1%7D%20%3D%201004.76%20%2B%200.5%2A1599.34%5C%5Cu_%7B1%7D%20%3D%201804.43%20kJ%2Fkg)
![s_{1} = s_{f1} + x_{1} * s_{fg1}\\s_{1} = 2.6456 + 0.5*3.5412\\s_{1} = 4.4162 kJ/kg](https://tex.z-dn.net/?f=s_%7B1%7D%20%3D%20s_%7Bf1%7D%20%2B%20x_%7B1%7D%20%2A%20s_%7Bfg1%7D%5C%5Cs_%7B1%7D%20%3D%202.6456%20%2B%200.5%2A3.5412%5C%5Cs_%7B1%7D%20%3D%204.4162%20kJ%2Fkg)
![Q_{12} = m(u_{2} - u_{1} ) + W_{12} \\600 = 0.5892(u_{2} -1804.43) + 124\\](https://tex.z-dn.net/?f=Q_%7B12%7D%20%3D%20m%28u_%7B2%7D%20-%20u_%7B1%7D%20%29%20%2B%20W_%7B12%7D%20%5C%5C600%20%3D%200.5892%28u_%7B2%7D%20-1804.43%29%20%2B%20124%5C%5C)
Solving for u₂
![u_{2} = 2612.3 kJ/kg](https://tex.z-dn.net/?f=u_%7B2%7D%20%3D%202612.3%20kJ%2Fkg)
Since P₂ = 1.2 MPa, u₂ = 2612.2 kJ/kg,
then from steam table, T₂ = 200°C, S₂ = 6.5898 kJ/kg-K
The entropy generated will be:
![\triangle S = m(S_{2} -S_{1} ) - \frac{Q_{12} }{T_{H} }\\ \triangle S= 0.5892(6.5898 - 4.4162) - \frac{600 }{573 }\\ \triangle S =0.233 kJ/K](https://tex.z-dn.net/?f=%5Ctriangle%20S%20%3D%20m%28S_%7B2%7D%20-S_%7B1%7D%20%29%20-%20%5Cfrac%7BQ_%7B12%7D%20%7D%7BT_%7BH%7D%20%7D%5C%5C%20%20%5Ctriangle%20S%3D%200.5892%286.5898%20-%204.4162%29%20-%20%5Cfrac%7B600%20%7D%7B573%20%7D%5C%5C%20%5Ctriangle%20S%20%3D0.233%20kJ%2FK)
Since ΔS > 0, this process is possible
Answer:
![\Delta V_{gas}=0.018m^3](https://tex.z-dn.net/?f=%5CDelta%20V_%7Bgas%7D%3D0.018m%5E3)
Explanation:
We define as,
c.V Piston
![(E_2-E_1)_{Pist}=m(u_2-u_1)+m[\frac{1}{2}V^2_2-0]+mg(h_2-0)](https://tex.z-dn.net/?f=%28E_2-E_1%29_%7BPist%7D%3Dm%28u_2-u_1%29%2Bm%5B%5Cfrac%7B1%7D%7B2%7DV%5E2_2-0%5D%2Bmg%28h_2-0%29)
![(E_2-E_1)_{Pist}=0+25*0.5*25^2+25*9.8*5](https://tex.z-dn.net/?f=%28E_2-E_1%29_%7BPist%7D%3D0%2B25%2A0.5%2A25%5E2%2B25%2A9.8%2A5)
![(E_2-E_1)_{Pist}=7712+5+1225.8=9038.3J](https://tex.z-dn.net/?f=%28E_2-E_1%29_%7BPist%7D%3D7712%2B5%2B1225.8%3D9038.3J)
The energy equation for the piston is
![E_2-E_1=W_{gas}-W_{atm}=P_{avg}\Delta V_{gas}-P_0 \Delta V_{gas}](https://tex.z-dn.net/?f=E_2-E_1%3DW_%7Bgas%7D-W_%7Batm%7D%3DP_%7Bavg%7D%5CDelta%20V_%7Bgas%7D-P_0%20%5CDelta%20V_%7Bgas%7D)
Remember that
![\Delta V_{atm}=-\Delta V_{gas}](https://tex.z-dn.net/?f=%5CDelta%20V_%7Batm%7D%3D-%5CDelta%20V_%7Bgas%7D)
So,
![\Delta V_{gas}=9.038kJ/(600-100)=0.018m^3](https://tex.z-dn.net/?f=%5CDelta%20V_%7Bgas%7D%3D9.038kJ%2F%28600-100%29%3D0.018m%5E3)
Explanation:
Artificial narrow intelligence is the type of Artificial Intelligence (AI) which can repeatedly perform tasks of limited scope.