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natulia [17]
3 years ago
8

If the average gauge pressure in the vein is 12200 Pa, what must be the minimum height of the bag in order to infuse glucose int

o the vein? Assume that the specific gravity of the solution is 1.02. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
Physics
1 answer:
Amanda [17]3 years ago
6 0

Answer:

h = 1.22 m

Explanation:

Given:

Pressure in the vein = 12200 Pa

Specific gravity of the liquid  = 1.02

now,

the pressure due to a fluid is given as:

P = pgh

where,

P is the pressure,

ρ is the density of fluid = specific gravity x density of water = 1.02 x 1000 kg/m³

ρ = 1020 kg/m³

g is the acceleration due to the gravity = 9.81m/s²

h is the height

thus,

h = P/pg =\frac{12200}{1020\times 9.8}=1.22 m

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We have that The ratio U1/U2 of their potential energies due to their interactions with Q is

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From the question we are told that

Question 1

Charge q1 is distance r from a positive point charge Q.

Question 2

Charge q2=q1/3 is distance 2r from Q.

Charge q1 is distance s from the negative plate of a parallel-plate capacitor.

Charge q2=q1/3 is distance 2s from the negative plate.

Generally the equation for the potential energy  is mathematically given as

U=\frac{-k*qQ}{r}

Therefore

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U1=\frac{-k*q_1Q}{r}

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U2=\frac{-k*q_1Q}{3*2r}

Since

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U1/U2=6

For Question 2

For U1

U1=\frac{-k*q_1Q}{s}\\\\For U2\\\\U2=\frac{-k*q_1Q}{3*2r}

Therefore

U1/U2=6

For more information on this visit

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7 0
3 years ago
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greater than

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All points to the left of zero on a horizontal number line are negative. A  number line<span> is a picture of a graduated straight </span>line<span> that serves as abstraction for real </span>numbers. <span>This </span>number line<span> can be extended to the left to </span>represent numbers<span> which are smaller than </span>0<span>. Such </span>numbers<span> are called negative </span>numbers<span>.</span>
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Under electrostatic conditions, the electric field just outside the surface of any charged conductor:
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Answer:

D. is always perpendicular to the surface of the conductor

Explanation:

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