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4 years ago

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Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 5.4×104 times the ac

celeration of gravity. The distance from the axis of rotation to the bottom of the test tube is 7.2 cm .

1 answer:

Luda [366]4 years ago

4 0

Answer: v= 195.2 m/s

Explanation:

The centripetal acceleration acts on the bottom of the test tube trying to take it closer to the axis of rotation, and it is defined by the following expression:

ac = v² / r

We know that ac = 5.4. 10⁴ . g = 5.4.10⁴.9.8 m/s²

Expressing r in m ⇒ r = 0.072 m.

So, the only unknown that remains is v, which is the linear speed of the bottom of the test tube, which we want to find out:

v = √ac. r = √5.4.10⁴.9.8 m/s². 0.072 m = 195.2 m/s

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A vertical plate has a sharp-edged orifice at its center. A water jet of speed V strikes the plate concentrically. Obtain an exp

Vesnalui [34]

Answer:

Force = 455672061 N

Explanation:

given data

V = 515 ft/s

D = 54 in = 4.5 ft

d = 51 in = 4.25 ft

to find out

Evaluate the force

solution

we know that force is express as

force F = ρ ΔA v² ..................1

put here value we get

force F = 1000 × $\frac{\pi}{4}$ [ D² - d² ] × v²

F = 1000 × $\frac{\pi}{4}$ [ 4.5² - 4.25² ] × 515²

Force = 455672061 N

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4 years ago

To reduce the global emission of greenhouse gases, which of the following would be theMOST effective and practical lifestyle cha

Anestetic [448]

Answer:

C. Decrease your consumption of beef.

Explanation:

The production of beef needs lots of land, which leads to trees being cut down, releasing carbon dioxide. Also, the ruminant animals used to produce beef end up farting a lot from their diets, and these farts are of high metane, which leads to greenhouse gases being emitted in the athmosphere.

Common plant proteins, otherwise, produce way less of these gases.

So the correct answer is:

C. Decrease your consumption of beef.

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3 years ago

What is structural analysis

Luden [163]

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3 years ago

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The pressure gage on a 2.5-m^3 oxygen tank reads 500 kPa. Determine the amount of oxygen in the tank if the temperature is 28°C

s2008m [1.1K]

Answer:

19063.6051 g

Explanation:

Pressure = Atmospheric pressure + Gauge Pressure

Atmospheric pressure = 97 kPa

Gauge pressure = 500 kPa

Total pressure = 500 + 97 kPa = 597 kPa

Also, P (kPa) = 1/101.325 P(atm)

Pressure = 5.89193 atm

Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)

Temperature = 28 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.15 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K

⇒n = 595.76 moles

Molar mass of oxygen gas = 31.9988 g/mol

Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g

7 0

3 years ago

Consider the following hypothetical scenario for Jordan Lake, NC. In a given year, the average watershed inflow to the lake is 9

dybincka [34]

Answer:

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

$V = V_{in}+V_{p}-V_{e}-V_{out}$ (Eq. 1)

Where:

$V$ - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

$V_{in}$ - Inflow amount of water, measured in cubic feet per year.

$V_{out}$ - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

$V_{p}$ - Amount of water due to precipitation, measured in cubic feet per year.

$V_{e}$ - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

$V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t$

$V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l}$ (Eq. 2)

Where:

$f_{in}$ - Average watershed inflow, measured in cubic feet per second.

$f_{out}$ - Average flow to be released, measured in cubic feet per second.

$\Delta t$ - Yearly time, measured in seconds per year.

$h_{p}$ - Change in lake height due to precipitation, measured in feet per year.

$h_{e}$ - Change in lake height due to evaporation, measured in feet per year.

$A_{l}$ - Surface area of the lake, measured in square feet.

If we know that $f_{in} = 900\,\frac{ft^{3}}{s}$ , $f_{out} = 300\,\frac{ft^{3}}{s}$ , $\Delta t = 31,536,000\,\frac{second}{yr}$ , $h_{p} = 32\,\frac{in}{yr}$ , $h_{e} = 55\,\frac{in}{yr}$ and $A_{l} = 47,000\,acres$ , the available amount of water for supply purposes in the Triangle area is:

$V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)$ $V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}$

The lake can withdraw a maximum of $1.464\times 10^{10}$ cubic feet per year to provide water supply for the Triangle area.

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3 years ago