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DENIUS [597]
3 years ago
7

What pressure, in millimeters of mercury (mm Hg), is equivalent to 2.13 atmospheres

Physics
1 answer:
Sati [7]3 years ago
6 0
The "Standard Atmosphere" is 760 mm Hg . 2.13 times that pressure is (2.13 x 760) = 1,618.8 mm Hg.
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2 years ago
Read 2 more answers
45.678 g to 4 digits
PolarNik [594]

Answer:

<h2>45.68 g</h2>

Explanation:

Given 45.678 g, converting the value to 4 digits is similar to converting the decimal number to 4 significant figure. We can see that the decimal number initially contains 5 numbers. To covert to a 4 digits number we will convert the last digit which is 8 to 0 because 0 is insignificant.

If the number we are converting to zero is more than 4 then the preceding number (7) will be rounded up by adding 1 to it

The decimal number will become 45.68.

Note that the value of 7 was rounded up to 8 since the value we are converting to zero is more than 4.

<em>Hence the decimal value in 4 digits is 45.68 g</em>

7 0
3 years ago
Consider a vortex filament of strength Γ in the shape of a closed circular loop of radius R. Consider also a straight line throu
Sedbober [7]

Answer:

\vec{V} = \frac{\Gamma}{2R}\vec{A}

Explanation:

We define our values according to the text,

R= Radius

\vec{V} =Velocity

\Gamma =Strenght of the vortex filament

From this and in a vectorial way we express an elemental lenght of this filmaent as \vec{dl}. So,

\vec{dl}x\vec{r} = R*dl*\vec{A}

Where \vec{A} imply a vector acting perpendicular to both vectors.

Applying Biot-Savart law, we have,

\vec{V} =\frac{\Gamma}{4\pi}\int\frac{\vec{dl}x\vec{r}}{r^3}

Substituting the preoviusly equation obtained,

\vec{V} = \frac{\Gamma}{4\pi}\int\frac{R*dl*\vec{A}}{R^3}

\vec{V} = \frac{\Gamma}{4\pi R^2}\int^{2\pi R}_0 dl*\vec{A}

\vec{V} = \frac{\Gamma(2\pi R \vec{A})}{4\pi R^2}

So we can express the velocity induced is,

\vec{V} = \frac{\Gamma}{2R}\vec{A}

6 0
3 years ago
If two coils placed next to one another have a mutual inductance of 3.00 mH, what voltage (in V) is induced in one when the 2.50
hodyreva [135]

Answer:

-0.1875 V.

Explanation:

Using

E₂ = MdI₁/dt........................ Equation 1

Where E₂ = Voltage induced in the second coil, M = mutual inductance of both coil, dI₁ = change in current in the first coil, dt = change in time.

Given: M = 3.00 mH = 0.003 H, dI₁ = (0-2.50) = -2.5 A, dt = 40 ms = 0.04 s.

Substitute into equation 1

E₂ = 0.003(-2.5)/0.04

E₂ = -0.1875 V.

Hence the induced emf = -0.1875 V.

3 0
3 years ago
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