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3 years ago

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Physics

1 answer:

xenn [34]3 years ago

5 0

Answer:

15.7079632679

Explanation:

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Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of

Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{high} = 15.999\,kPa$

$p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{low} = 10.666\,kPa$

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}$

$p_{high} = 2.320\,psi$

$p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}$

$p_{low} = 1.547\,psi$

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

$\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}$

$h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}$ (Eq. 1)

Where:

$\rho_{w}$ , $\rho_{Hg}$ - Densities of water and mercury, measured in kilograms per cubic meter.

$h_{w}$ , $h_{Hg}$ - Heights of water and mercury columns, measured in meters.

If we know that $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , $h_{Hg, high} = 0.120\,m$ and $h_{Hg, low} = 0.080\,m$ , then we get that:

$h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m$

$h_{w, high} = 1.632\,m$

$h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m$

$h_{w, low} = 1.088\,m$

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

4 0

3 years ago

10. How do you use a triple-beam balance? Fill in the blanks.

wolverine [178]

Answer:

1st – Place the film canister on the <u>scale</u>.

2nd – Slide the large <u>weight </u>to the right until the arm drops below the line and then move it back one notch.

3rd – Repeat this process with the <u>top</u> weight. When the arm moves below the line, back it up one groove.

4th – Slide the <u>small </u>weight on the front beam until the <u>lines</u> match up.

5th – Add the amounts on each beam to find the total <u>mass </u>to the nearest tenth of a gram.

Explanation:

The triple beam balance is an instrument that is used in measuring the mass of substances to a very high degree of precision. The reading error is given by ±0.05 grams. The triple beam balance as the name implies has three beams that measure substances of different mass levels.

The beams are categorized as small, medium, and large. There is a balance on which the substance to be weighed is placed directly upon. To use this measuring device, the procedures mentioned above are followed.

5 0

3 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

8 0

3 years ago

The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$