Answer:
density of water is much greater than density of air
Explanation:
when body is in fluid it carries weight which is always equal to upthrust
upthrust is given as
upthrust= density of fluid ×strength of gravitational field g × volume of body in fluid
as volume of ship and aeroplane is same and strength of gravitational field or acceleration due to gravity is same hence upthrust depens upon density of fluid .
density of air is about 1.2 kg/m^3 and density of water is 1000kg/m^3 hence ship carries more load
Answer: radio and television
Satellite orbits with an East-West orientation are best used
for communication purposes. The satellite orbit should always be suited for its
purpose. Radio and television are powered by numerous satellites rotating the
Earth from east to west.
Potential Energy is calculated through the equation,
PE = m x g x h
where m is the mass, g is the acceleration due to gravity, and h is the height. Since, weight is equal to mass times gravity, PE may also be calculated through the equation,
PE = W x h
Substituting the known values,
PE = (2,200 N) x (15 m)
<em>PE = 33,000 J</em>
The question is incomplete. The complete question is :
The pressure difference, Δp, ac
ross a partial blockage in an artery (called a stenosis) is approximated by the equation :
![$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$](https://tex.z-dn.net/?f=%24%5CDelta%20p%3DK_v%5Cfrac%7B%5Cmu%20V%7D%7BD%7D%2BK_u%5Cleft%28%5Cfrac%7BA_0%7D%7BA_1%7D-1%5Cright%29%5E2%20%5Crho%20V%5E2%24)
Where V is the blood velocity, μ the blood viscosity {FT/L2}, ρ the blood density {M/L3}, D the artery diameter,
the area of the unobstructed artery, and A1 the area of the stenosis. Determine the dimensions of the constants
and
. Would this equation be valid in any system of units?
Solution :
From the dimension homogeneity, we require :
![$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$](https://tex.z-dn.net/?f=%24%5CDelta%20p%3DK_v%5Cfrac%7B%5Cmu%20V%7D%7BD%7D%2BK_u%5Cleft%28%5Cfrac%7BA_0%7D%7BA_1%7D-1%5Cright%29%5E2%20%5Crho%20V%5E2%24)
Here, x means dimension of x. i.e.
![$[ML^{-1}T^{-2}]=\frac{[K_v][ML^{-1}T^{-1}][LT^{-1}]}{[L]}+[K_u][1][ML^{-3}][L^2T^{-2}]$](https://tex.z-dn.net/?f=%24%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D%3D%5Cfrac%7B%5BK_v%5D%5BML%5E%7B-1%7DT%5E%7B-1%7D%5D%5BLT%5E%7B-1%7D%5D%7D%7B%5BL%5D%7D%2B%5BK_u%5D%5B1%5D%5BML%5E%7B-3%7D%5D%5BL%5E2T%5E%7B-2%7D%5D%24)
![$=[K_v][ML^{-1}T^{-2}]+[K_u][ML^{-1}T^{-2}]$](https://tex.z-dn.net/?f=%24%3D%5BK_v%5D%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D%2B%5BK_u%5D%5BML%5E%7B-1%7DT%5E%7B-2%7D%5D%24)
So,
dimensionless
So,
and
are dimensionless constants.
This equation will be working in any system of units. The constants
and
will be different for different system of units.