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Maslowich
3 years ago
7

Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S ​

Physics
1 answer:
Tems11 [23]3 years ago
6 0

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

v_{f}=v_{o}+a*t

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}

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The pressure difference, , across a partial blockage in an artery (called a stenosis) is approximated by the equation where is t
Strike441 [17]

The question is incomplete. The complete question is  :

The pressure difference, Δp, acK_uross a partial blockage in an artery (called a stenosis) is approximated by the equation :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Where V is the blood velocity, μ the blood viscosity {FT/L2}, ρ the blood density {M/L3}, D the  artery diameter, A_0 the area of the unobstructed artery, and A1 the area of the stenosis.  Determine the dimensions of the constants K_v and K_u. Would this equation be valid in any  system of units?

Solution :

From the dimension homogeneity, we require :

$\Delta p=K_v\frac{\mu V}{D}+K_u\left(\frac{A_0}{A_1}-1\right)^2 \rho V^2$

Here, x means dimension of x. i.e.

$[ML^{-1}T^{-2}]=\frac{[K_v][ML^{-1}T^{-1}][LT^{-1}]}{[L]}+[K_u][1][ML^{-3}][L^2T^{-2}]$

                    $=[K_v][ML^{-1}T^{-2}]+[K_u][ML^{-1}T^{-2}]$

So, $[K_u]=[K_v]=[1 ]=$ dimensionless

So, K_u and K_v  are dimensionless constants.

This equation will be working in any system of units. The constants K_u and K_v will be different for different system of units.

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A 2.0 kg mass weighs 10 Newtons on planet X. what is the acceleration due to gravity on planet X? Show the work.
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\frac{F_E}{g_E}=\frac{F_X}{g_X} \implies\\g_x = \frac{g_E\cdot F_X}{F_E}=\frac{9.8 N/kg \cdot 10 N }{19.6N}=5 \frac{N}{kg}

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Explanation:

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4 years ago
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