Answer:
Explanation:
A )
When empty , H₀ length of barge is inside water .
volume of barge inside water = A x H₀
Weight of displaced water = AH₀ x ρ x g
Buoyant force = weight of displaced water = AH₀ ρg
B)
It should balance the weight of barge
Weight = buoyant force
Weight = AH₀ ρg
mass of barge = weight / g
weight / g = AH₀ ρ
= 550 x .55 x 1000
= 302500 kg
Dry ice is carbon dioxide is classified as a compound <span>and its physical state solid. A compound is any substance that consists of two or more seperate elements. The elements in this case are carbon and oxygen. Dry ice finds application in many different industries such as in the production of ice cream. </span>
Answer:
high accuracy low precision.
Explanation:
Answer:
Cold Front 3 // Stationary Front 1 // Warm Front 2 // Occluded Front 4
Explanation:
It's simple. Warm front means the warm air is pressing forward, which is why it's a warm front. Stationary Front, meaning they're at a standstill, also makes sense because stationary means not moving. Then since your last option is Occluded Front, since the others already have an answer, you have no choice but to match 4 with it. I took the quiz and got the answer right. :D
Answer:
Part a)
![\rho = 0.0205 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%200.0205%20kg%2Fm%5E3)
Part b)
![\rho = 1.22 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%201.22%20kg%2Fm%5E3)
So density of atmosphere at Martian Surface is very less than the density at Earth.
Explanation:
Part a)
As per ideal gas equation we know that
![PM = \rho RT](https://tex.z-dn.net/?f=PM%20%3D%20%5Crho%20RT)
here we know that Martian atmosphere is equivalent to that of carbon
so we will have
![P = 900 Pa](https://tex.z-dn.net/?f=P%20%3D%20900%20Pa)
![T = 273 - 41 = 232 K](https://tex.z-dn.net/?f=T%20%3D%20273%20-%2041%20%3D%20232%20K)
now we will have
![(900)(0.044) = \rho (8.31)(232)](https://tex.z-dn.net/?f=%28900%29%280.044%29%20%3D%20%5Crho%20%288.31%29%28232%29)
![\rho = 0.0205 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%200.0205%20kg%2Fm%5E3)
Part b)
Now for the earth surface the density of air is given for
![P = 101.6 kPa](https://tex.z-dn.net/?f=P%20%3D%20101.6%20kPa)
![T = 18 ^oC](https://tex.z-dn.net/?f=T%20%3D%2018%20%5EoC)
so we will have
![PM = \rho RT](https://tex.z-dn.net/?f=PM%20%3D%20%5Crho%20RT)
![(101.6\times 10^3)(0.029) = \rho(8.31)(273 + 18)](https://tex.z-dn.net/?f=%28101.6%5Ctimes%2010%5E3%29%280.029%29%20%3D%20%5Crho%288.31%29%28273%20%2B%2018%29)
![\rho = 1.22 kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%201.22%20kg%2Fm%5E3)
So density of atmosphere at Martian Surface is very less than the density at Earth.