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.
Consider a car<span> that travels between points A and B. The </span>car's<span> average </span>speed<span> can be ..... the </span>car<span> to </span>slow down<span> with a </span>constant acceleration<span> of </span>magnitude 3.50 m/s2<span>. </span>If<span> the </span>car comes<span> to a </span>stop<span> in a </span>distance<span> of</span>30.0 m<span>, what was the </span>car's original speed<span>? ... A </span>car<span> is </span>traveling<span> at 26.0 </span>m<span>/s when the </span>driver suddenly applies<span> the </span>brakes<span>, ...</span>
A concave lens can only form a virtual image. The correct option among all the options that are given in the question is the third option or option "C". Concave lenses are mostly thinner in the middle compared to its edges. I hope that this answer has come to your help.
Given Information:
Length of wire = 132 cm = 1.32 m
Magnetic field = B = 1 T
Current = 2.2 A
Required Information:
(a) Torque = τ = ?
(b) Number of turns = N = ?
Answer:
(a) Torque = 0.305 N.m
(b) Number of turns = 1
Explanation:
(a) The current carrying circular loop of wire will experience a torque given by
τ = NIABsin(θ) eq. 1
Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.
We know that area of circular loop is given by
A = πr²
where radius can be written as
r = L/2πN
So the area becomes
A = π(L/2πN)²
A = πL²/4π²N²
A = L²/4πN²
Substitute A into eq. 1
τ = NI(L²/4πN²)Bsin(θ)
τ = IL²Bsin(θ)/4πN
The maximum toque occurs when θ is 90°
τ = IL²Bsin(90)/4πN
τ = IL²B/4πN
torque will be maximum for N = 1
τ = (2.2*1.32²*1)/4π*1
τ = 0.305 N.m
(b) The required number of turns for maximum torque is
N = IL²B/4πτ
N = 2.2*1.32²*1)/4π*0.305
N = 1 turn
Answer:
65.87 s
Explanation:
For the first time,
Applying
v² = u²+2as.............. Equation 1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance
From the question,
Given: u = 0 m/s (from rest), a = 1.99 m/s², s = 60 m
Substitute these values into equation 1
v² = 0²+2(1.99)(60)
v² = 238.8
v = √238.8
v = 15.45 m/s
Therefore, time taken for the first 60 m is
t = (v-u)/a............ Equation 2
t = (15.45-0)/1.99
t = 7.77 s
For the final 40 meter,
t = (v-u)/a
Given: v = 0 m/s(decelerates), u = 15.45 m/s, a = -0.266 m/s²
Substitute into the equation above
t = (0-15.45)/-0.266
t = 58.1 seconds
Hence total time taken to cover the distance
T = 7.77+58.1
T = 65.87 s
