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 :
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 :
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, ![$[K_u]=[K_v]=[1 ]=$](https://tex.z-dn.net/?f=%24%5BK_u%5D%3D%5BK_v%5D%3D%5B1%20%5D%3D%24)
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.
Answer:
t=1.4hours
Explanation:
The half life is 1hour
At t=0 he has a mass of 4kg
So he want it to be 1kg, so that his weapon can work.
Applying the exponential function of decay
M=Cexp(-kt)
Where,
M is the mass at any time
C is a constant of integration
k is the rate of decay
Given that it has an half life of 1 hours.
Then k is 1
At t =0 the mass is 4kg
Therefore
4=Cexp(0)
C=4
M=4exp(-kt)
Since rate of decay is 1, then k=1
M=4exp(-t)
We need to find t at M=1kg
1=4exp(-t)
1=4exp(-t)
1/4=exp(-t)
0.25=exp(-t)
Take In of both sides
In(0.25)=-t
-1.3863=-t
Then, t=1.386hour
Then it will take about 1.4 hours to get to 1kg.
Answer:
The force the rock exerts on Sarah =#is 65 N
Explanation:
The given parameters are;
Sarah's weight = 392 N
The force with which Sarah pushes the rock = 65 N
The mass of the rock = 56 kg
The weight of the rock = The mass of the rock × Acceleration due to gravity
∴ The weight of the rock = 56 kg ×9.81 m/s² = 549.36 N
Given that the force Sarah applies to push the rock = 65 N, then by Newton's third law of motion which states that action and reaction are equal and opposite, the force that the rock exert on Sarah is equal an opposite to the force Sarah is applying
Therefore, the force the rock exerts on Sarah = 65 N (in the opposite direction).
Answer:
AS- X 3.42 Y 3. B) X Y c) x Ross TET V V. a 1.71 1.71 LLL LLL 2.42 N al
Explanation:
Answer:
Explanation:
The rate of volume flow out of tank can be expressed as:
where,
dV/dt = Volume flow rate
A = Cross-sectional area of outlet = πd²/4
d = diameter of circular outlet
dL = Displacement covered by water
dt = time taken
but we know that:
Velocity = υ = displacement/time = dL/dt
Substituting the values of "dL/dt" and "A" in the equation, we get:
This is the expression for volume flow rate dV/dt, on terms pf v, d.
