24.11.2021

A typical laboratory centrifuge rotates at 4000 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation? Express your answer in meters per second squared. \[a_{1}=1.8 \times 10^{4} \mathrm{~m} / \mathrm{s}^{2}\] Previous Answers Correct Part B For comparison, what is the magnitude of the acceleration a test tube would experience if stopped in a \(1.0-\mathrm{ms}\)-long encounter with a hard floor after falling from height 1.0 m ? Express your answer in meters per second squared.

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To solve this problem, we will break it down into two parts, as indicated. ### Part A: Acceleration in the Centrifuge **Given:** - The centrifuge rotates at 4000 revolutions per minute (rpm). - The distance of the test tube from the axis of rotation is 10 cm, which we convert to meters: \( r = 10 \, \text{cm} = 0.10 \, \text{m} \). **Step 1: Convert RPM to Radians per Second** To find the acceleration, we first need to convert the rotation speed from revolutions per minute to radians per second. 1 revolution is equal to \( 2\pi \) radians. Thus: \[ \omega = 4000 \, \text{rpm} \times \frac{2\pi \, \text{radians}}{1 \, \text{revolution}} \times \frac{1 \, \text{minute}}{60 \, \text{seconds}} \] Calculating this gives: \[ \omega = 4000 \times \frac{2\pi}{60} = \frac{8000\pi}{60} \approx 418.88 \, \text{radians/second} \] **Step 2: Calculate the Acceleration** Centripetal acceleration (\( a \)) at a distance \( r \) from the axis of rotation is given by the formula: \[ a = r \omega^2 \] Substituting the values: \[ a = 0.10 \, \text{m} \times (418.88 \, \text{radians/second})^2 \] \[ a \approx 0.10 \times 175,157.56 \approx 17,515.76 \, \text{m/s}^2 \] When rounded, this gives: \[ a \approx 1.75 \times 10^{4} \, \text{m/s}^2 \] ### Part B: Acceleration Upon Stopping after Falling **Given:** - Height of the fall: \( h = 1.0 \, \text{m} \) - Duration of stopping (encounter with the floor): \( t = 1.0 \, \text{ms} = 0.001 \, \text{s} \) **Step 1: Calculate the Speed Just Before Impact** To find the speed just before impact, we can use the following kinematic equation: \[ v^2 = u^2 + 2gh \] where \( u = 0 \, \text{m/s} \) (initial velocity), \( g \approx 9.81 \, \text{m/s}^2 \), and \( h = 1 \, \text{m} \). Calculating: \[ v^2 = 0 + 2 \times 9.81 \times 1 \approx 19.62 \] \[ v \approx \sqrt{19.62} \approx 4.43 \, \text{m/s} \] **Step 2: Calculate the Magnitude of the Deceleration** When the test tube hits the floor, it comes to a stop, which means its final speed \( v_f = 0 \, \text{m/s} \). The magnitude of the deceleration (\( a \)) can be calculated using the formula: \[ a = \frac{v_f - v}{t} \] Substituting: \[ a = \frac{0 - 4.43 \, \text{m/s}}{0.001 \, \text{s}} = -4430 \, \text{m/s}^2 \] The negative sign indicates deceleration (i.e., slowing down). Therefore, the magnitude of the acceleration (deceleration) is: \[ |a| \approx 4430 \, \text{m/s}^2 \] ### Final Answers: - Part A: The acceleration at the end of a test tube in the centrifuge is approximately \( 1.75 \times 10^{4} \, \text{m/s}^2 \). - Part B: The magnitude of the acceleration when the test tube encounters the floor is \( 4430 \, \text{m/s}^2 \).
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1.a_{rad}=17545.2\frac{m}{s^{2}}

2.a=4429.45 \frac{m}{s^{2}}

Explanation:

Radial acceleration is:

a_{rad}=\frac{v^2}{r} (1)

With r the radius respects the axis of rotation and v the tangential velocity that is related with angular velocity (ω) by:

v= \omega r (2)

By (2) on (1):

a_{rad}= \frac{(\omega r)^2}{r}= (\omega )^2r=(418,88\frac{rad}{s})^2(0.1m)

a_{rad}=17545.2\frac{m}{s^{2}}

To find the acceleration of the tube with the fall, we can use the expression:

\overrightarrow{J}=\overrightarrow{F}_{avg}(\varDelta t) (3)

Due impulse-momentum theorem:

\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i} (4)

with p the momentum and J the impulse. By (4) on (3):

\overrightarrow{p}_{f}-\overrightarrow{p}_{i}=\overrightarrow{F}_{avg}(\varDelta t)

And using Newton's second law (F=ma) and that (P=mv):

mv_f-mv_i=(ma)(\varDelta t) (5)

Final velocity is the velocity just after the encounter with hard floor, and initial momentum us just before that moment so the first one is zero and the second one can be found sing conservation of energy:

\frac{mv_i}{2}=mgh

v_i=\sqrt{2gh}=\sqrt{2(9.81)(1.0)}=4.43\frac{m}{s}

So (5) is:

-m(4.43)=(ma)(\varDelta t)

solving for a:

a=\frac{4.43}{\varDelta t}=\frac{4.43}{1.0\times10^{-3}}=-4429.45

It’s negative because is opposed to the tube movement.

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The idea of profit and loss in grammar is known as a noun phrase. It functions as the subject or the object of a sentence, depending on the context.

In the sentence "Profit and loss are important concepts in business", the noun phrase "profit and loss" functions as the subject. It identifies what is important.

In the sentence "The company experienced significant profit and loss last year", the noun phrase "profit and loss" functions as the object. It receives the action of the verb "experienced" and specifies what the company experienced.

Noun phrases like "profit and loss" can be made up of multiple nouns or noun equivalents and can function in various ways within a sentence.
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The type of language used in the question "Are there any reasons why this idea would not work?" is speculative.
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To solve this problem, first we need to understand that the volume of punch stays the same (2 gallons) before and after Gracie makes her adjustments. Therefore, we can create an equation using the information we have: the original amount of juice, how much juice Gracie adds, and the final amount of juice in the punch.

Let X = the amount of punch that Gracie pours out (and consequently, the volume of 100% juice she adds).

The initial punch is comprised of 50% (or 0.5) juice, therefore, the volume of the juice in the initial punch would be 0.5 * 2 = 1 gallon.

If Gracie pours out X gallons of punch, she also takes out 0.5 * X gallons of juice (because the original punch is 50% juice).

The final punch which is 2 gallons is comprised of 65% (or 0.65) juice, therefore, the volume of the juice in the final punch would be 0.65 * 2 = 1.3 gallons.

According to the problem, the final volume of juice (1.3 gallons) is equal to the initial volume of juice (1 gallon) minus the juice Gracie removed (0.5 * X) plus what she replaced it with (X gallons of 100% juice).

Therefore, we can write the equation:

1 - 0.5*X + X = 1.3

Solving this equation, we collect like terms and get:

0.5*X = 0.3
Which simplifies to:
X = 0.3/0.5 = 0.6 gallons

So, Gracie poured out 0.6 gallons of her original punch mixture and replaced it with pure 100% juice to achieve the 2 gallons of punch that contains 65% juice.


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To solve this problem, you need to know that the external angle of a triangle is equal to the sum of the two non-adjacent angles.

Here, ∠MLK is the external angle of ∆ZML.

So m∠MLK = m∠ZLK + m∠MLZ, which translates into:

130 = (x + 86) + (x + 66)

Combine the constants on the right hand side:

130 = 2x + 152

To isolate 2x, subtract 152 from both sides:

130 - 152 = 2x

-22 = 2x

Now divide each side by 2 to solve for x:

x = -22 / 2

x = -11

Now plug x = -11 into m∠ZLK = x + 86 to find m∠ZLK:

m∠ZLK = -11 + 86

m∠ZLK = 75

Therefore, m∠ZLK = 75°.
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To determine the pressure exerted by the trapped air in the mercury barometer, we can use the concept of equilibrium.

First, let's understand the setup. The mercury barometer consists of a long, vertical tube filled with mercury, inverted in a dish containing mercury. The top of the tube is sealed, creating a vacuum above the mercury column. The height of the mercury column indicates the atmospheric pressure.

When there is trapped air inside the tube, it exerts a pressure on the mercury column, balancing the atmospheric pressure.

To solve the problem, we need to compare the height of the mercury column in the barometer to that of the aneroid barometer, which reads 75 cm Hg. This means that the pressure exerted by the trapped air is equivalent to the pressure of a mercury column measuring 75 cm.

To calculate the pressure exerted by the trapped air, we use the equation for hydrostatic pressure:

Pressure = Density × Gravity × Height

In this case, the density will be that of mercury (which we'll denote as ρ) and the height of the mercury column will be 75 cm.

The equation becomes:

Pressure = ρ × g × Height

Now, let's substitute the values for density and gravity:

Pressure = (13.6 g/cm³) × (9.8 m/s²) × (75 cm)

Since the density is given in grams per cubic centimeter (g/cm³) and gravity is given in meters per second squared (m/s²), we need to convert the density to kilograms per cubic meter (kg/m³) to maintain consistency in the units. 1 g/cm³ is equal to 1000 kg/m³.

Substituting the values and performing the necessary conversions, we get:

Pressure = (13.6 × 1000) kg/m³ × 9.8 m/s² × (0.75 m)

Pressure = 127,680 kg/(m·s²) × 0.75 m

Pressure = 95,760 Pa

Therefore, the pressure exerted by the trapped air in the mercury barometer is approximately 95,760 Pascal (Pa).

Note: Pascal (Pa) is the unit of pressure in the International System of Units (SI).
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B. to help keep track of the developments in the meeting and any future action items to encourage all the participants contribute during the meeting.

Taking minutes during a meeting is important for several reasons:

1. Documentation: Minutes serve as a written record of what was discussed, decisions made, and actions planned during a meeting. They provide a historical reference that can be used for future reference, clarification, or legal purposes. Having an accurate record helps avoid any misunderstandings or disagreements about what was agreed upon.

2. Accountability: By assigning someone to take minutes, it ensures that there is a designated individual responsible for capturing the key points and outcomes of the meeting. This brings accountability and helps ensure that nothing important is missed or overlooked.

3. Tracking progress: Minutes help keep track of the developments and progress made during the meeting. They outline action items, responsibilities, and deadlines, making it easier to follow up on tasks and evaluate the progress at subsequent meetings. Without minutes, it can be challenging to remember all the details and actions agreed upon, leading to potential inefficiencies and missed opportunities.

4. Participation: Knowing that minutes will be recorded, participants are more likely to actively contribute during the meeting. They recognize that their ideas, suggestions, and feedback will be documented and taken into consideration. This promotes active engagement and collaboration among participants, leading to more productive meetings.

While option A and D may be part of the leader's responsibilities, they are not the primary reasons for taking minutes. Minutes primarily serve to keep a written record and facilitate the tracking of developments and action items during the meeting.
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The constraints are as follows based on the given conditions:

1. x + y ≤ 200 (The total number of plants Lee grows should be less than or equal to 200.)
2. 30 ≤ x ≤ 120 (Lee wants to grow at least 30 but no more than 120 basil plants.)
3. y ≥ 50 (Lee wants to grow at least 50 parsley plants.)

From these, we can see that the condition x ≥ 30 does not imply any limitations for this problem. This is because the constraint requiring at least 30 and no more than 120 basil plants has already been expressed as 30 ≤ x ≤ 120. Any additional stipulation of x ≥ 30 is redundant and does not place any new restrictions on this problem. Therefore, x ≥ 30 is not a constraint for this problem.
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The objective function in this problem is the function that represents the total revenue from selling banana and strawberry shakes, which needs to be maximized.

The revenue from selling x banana shakes is 3.25x (since each banana shake sells for $3.25).

The revenue from selling y strawberry shakes is 2.75y (since each strawberry shake sells for $2.75).

Therefore, to maximize their revenue, the juice bar owners would like to maximize the function 3.25x + 2.75y.

So, the objective function for this problem is:

Z = 3.25x + 2.75y

Simply put, the objective function is the equation in a linear programming problem that is either maximized or minimized. In this case, the owners of the juice bar are looking to maximize their revenue (Z) by selling banana shakes (represented by x) and strawberry shakes (represented by y). The coefficients 3.25 and 2.75 in front of the variables x and y represent the prices of the banana and strawberry shakes, respectively; multiplying these coefficients with the variables gives the total revenue from selling the respective shakes. Through this function, the owners can find out how many of each type of shakes they need to sell to maximize their revenue given the constraints.

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