Answer:
Please select the correct statements concerning use of a centrifuge. (Select all that apply.)
(a) When placing test tubes in the centrifuge, what are the best practices?
- Wait until all positions on the rotor contain a test tube before activating the centrifuge.
- <u>Place test tubes containing approximately equal volumes in opposite positions on the rotor.</u>
- Carefully stopper the test tubes to ensure the liquids do not spill out.
- <u>Never put stoppers or other items in the centrifuge with your tubes</u>.
(b) When operating the centrifuge, students should do which of the following?
- <u>Make sure that the lid is completely shut and the safety knob is turned to engage the safety switch.</u>
- <u>Centrifuge for approximately 30 minutes</u>.
- Centrifuge for approximately 30 seconds.
- Mid-way through the process, open the lid of the centrifuge and check to be sure it is operating.
(c) When the centrifuge process is complete, students should do which of the following?
- <u>Using the eye-hole on top, monitor the spinning rate and open the lid only after the rotor has slowed significantly</u>.
- Immediately open the lid of the centrifuge to remove the test tubes before they are damaged.
- <u>Wait for the rotor to come to a complete stop before attempting to remove your test tubes</u>.
- Reach out and carefully slow the rotor with your hand so you may finish the lab faster.
Explanation:
I UNDERLINED and BOLD the answers that apply
Answer:
<em>a) 42 mm</em>
<em>b) 144.4 MPa</em>
<em></em>
Explanation:
Load P = 200 kN = 200 x 10^3 N
Torque T = 1.5 kN-m = 1.5 x 10^3 N-m
maximum shear stress τ = 100 Mpa = 100 x 10^6 Pa
diameter of shaft d = ?
From T = τ * *
substituting values, we have
1.5 x 10^3 = 100 x 10^6 x x
= 7.638 x 10^-5
d = = 0.042 m = <em>42 mm</em>
b) Normal stress = P/A
where A is the area
A = = = 1.385 x 10^-3
Normal stress = (200 x 10^3)/(1.385 x 10^-3) = 144.4 x 10^6 Pa = <em>144.4 MPa</em>
You have to show the story or experiment so we know
Based on the calculations, the magnitude (a) of it's total acceleration is equal to 2.71 m/s².
<u>Given the following data:</u>
- Angle of inclination = 10°.
- Radius of curvature, r = 40 meters.
- Acceleration of the minivan, A = 1.8 m/s².
- Initial velocity, u = 0 m/s (since it's starting from rest).
<h3>How to determine the magnitude (a) of it's total acceleration?</h3>
First of all, we would determine the final velocity of the minivan by applying the first equation of motion as follows:
V = u + at
V = 0 + 1.8 × 5
V = 9 m/s.
Next, we would calculate the centripetal acceleration of this minivan:
Ac = V²/r
Ac = 9²/40
Ac = 2.025 m/s².
Now, we can determine the magnitude (a) of it's total acceleration:
a = √(Ac² + A²)
a = √(2.025² + 1.8²)
a = 2.71 m/s².
Read more on acceleration here: brainly.com/question/24728358
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Answer:
P = 0.490 kip
Explanation:
given data
allowable bearing stress = 2 ksi
allowable tensile stress = 18 ksi
diameter = 0.31 in
outer diameter = 0.75 in
inner diameter (hole) = 0.50 in
solution
we find here cross section area of shank that is express as
Area = ..................1
area =
area = 0.0754 in²
and
now we get here allowable load in bolt will be
...................2
P =
P = 18 × 10³ × 0.0754
P = 1357.2 = 1.357 kip
and
now find here area of washer is
Area = .......................3
put here value
Area =
area = 0.2454 in²
so now we get here allowable load of washer will be
.....................4
P = 2 × 10³ × 0.245
P = 490 = 0.490 kip