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Ivenika [448]
3 years ago
5

When drawing up liquid into a micropipette, put the tip in the liquid at a Choose... degree angle, push the plunger to the Choos

e... point of resistance, and then Choose... the plunger. When dispensing that liquid from the micropipette, hold the micropipette at a Choose... degree angle in the desired container and push the plunger to the Choose... point of resistance. Choose... the plunger until dispensing is done.
Physics
1 answer:
tangare [24]3 years ago
3 0

Answer:

Explanation:

This question seeks to test the knowledge of operating a micropipette.

When trying to draw a liquid from a container using a micropipette (tip), the micropipette should be <u>held vertically straight at an angle of 90 degrees</u> with the plunger pushed to the <u>first resistance</u> and then released.

When trying to dispense the drawn liquid into another container, the micropipette is <u>held vertically straight at the same 90 degrees</u> but the plunger is pushed to the <u>second/last</u> point of resistance. <u>You keep pressing</u> the plunger at the same degree and resistance until dispensing is done.

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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3 years ago
Which situation is the best example of translational motion?.
AleksAgata [21]

Answer:

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8 0
2 years ago
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Alex

Answer:

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EM waves have both Electrical and magnetic features.

they travel in the velocity of light (3*10⁸ ms⁻¹)

Electromagnetic spectrum is obtained according to their wave length and the frequency. Due to wave length range it's categorized. Here is the decreasing  order of wave length and increasing order  of different wave types in electromagnetic spectrum

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5 0
3 years ago
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Travka [436]
1) In a circular motion, the angular displacement \theta is given by
\theta =  \frac{S}{r}
where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire: S=2600 m. Using the information about the radius, r=0.35 m, we find the total angular displacement:
\theta =  \frac{2600 m}{0.35 m} =7428 rad

2) If we put larger tires, with radius r=0.60 m, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
\theta =  \frac{2600 m}{0.60 m}=4333 rad
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What is the kinetic energy f a 25kg object movingat a velocity of 10m/s?
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Using K.E=1/2MV^2
answer is 125joules
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