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2 years ago

10

Unele rigle gradate au partea pe care sunt imprimante graditatiile in forma de pana. prezinta aceasta vreo importanta pentru mas

urarea cu aceste rigle ?

1 answer:

GrogVix [38]2 years ago

7 0

Da ne ajuta pentru a putea citi corect lungimea ... coronită???

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A vacuum gage attached to a power plant condenser gives a reading of 27.86 in. of mercury. The surrounding atmospheric pressure

Nadusha1986 [10]

Answer:

absolute pressure = 1.07 lbft/in^2

Explanation:

given data:

vaccum gauge reading h = 27.86 inch = 2.32 ft

we know that

gauge pressure p is given as

$p = \rho gh$

$p = 848 \ lb/ft^3 * 32 \ ft/s^2 *2.32 ft = 62955.52 \ lbft/s^2 * 1/ft^2$

we know that $1\ lb ft\s^2 = \frac{1}{32.174}\ lbft$

1 ft = 12 inch

therefore $p = 62955.52 * \frac{1}{32.174} * \frac{1}{12^2}\ lbf/in^2$

$p = 13.59\ lbft/in^2$

$P_{atm} = 14.66\ lbf/in^2$

so absolute pressure = $P_{atm} - p$

= 14.66 - 13.59 = 1.07 lbft/in^2

4 0

2 years ago

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, ac $K_u$ ross a partial blockage in an artery (called a stenosis) is approximated by the equation :

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

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.

5 0

3 years ago

What is the primary mechanical advantage of a screw?

9966 [12]

What well u can use to make a shelter but that's all I can think of ??

5 0

3 years ago

Read 2 more answers

Unit Test Review

PolarNik [594]

Answer:

option B...

they represent different concept...

i hope this will answer your question

4 0

1 year ago

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acce

Lana71 [14]

Answer:

Explanation:

Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero

F - Force

T = Tension

m = mass

a = acceleration

g = gravitational force

Let the given Normal on block 2 = N

and $N = m_2 g \cos \theta$

and the tension in the given string is said to be $T = m_2 g \sin \theta$

When the acceleration $a=\frac{F}{m_1}$

for the said block 1.

It will definite be zero only when Force is zero , F=0.

Here by Force, F

I refer net force on block 1.

Now we know

$F = m_1g-T.$

It is known that if the said

$\theta=\frac{\pi}{2}$ ,

then Tension $T= m_2g$ $[since \sin(\pi/2) = 1]$ ,

Now making $"F = m_1g - m_2g"$

So If we are to make Force equal to zero

$F=0 => m_1g = m_2g \ or \ m_1 = m_2$

6 0

3 years ago