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USPshnik [31]
3 years ago
6

which part of the microscope will be used first to adjust the focus when starting with the lowest power lens?

Physics
1 answer:
noname [10]3 years ago
7 0

The first part of the microscope that should first be used to adjust the focus when starting with the lowest power lens would be the coarse adjustment knob.

There are two knobs in a typical light microscope with which objects on slides can be brought into focus:

  1. Coarse adjustment knob
  2. Fine adjustment knob

The 2 knobs are used to adjust the stage to either bring it up towards the objective lens or down away from them. The coarse adjustment knob, however, moves the stage a considerable distance with each turn. The fine adjustment knob, on the other hand, only moves the stage very little with each turn.

The lowest power lenses are often short. Hence, using the coarse adjustment knob is ideal in order to quickly bring objects on slides into focus.

The fine adjustment knob comes highly recommended at high objectives because high objectives lenses are usually long and using the coarse adjustment knob can lead to a breakage of the slide by the lens.

More on bringing objects into focus on a microscope can be found here: brainly.com/question/24319677

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A cylinder contains 3.0 L of oxygen at 310 K and 2.5 atm. The gas is heated, causing a piston in the cylinder to move outward. T
Alex_Xolod [135]

Answer:

The gas pressure is: 1.55 atm.

Explanation:

We need to use the equation that relate the variables given at the exercise (pressure, temperature and volume) from the ideal gas law formula, when the mass is constant we can reduce the expretion PV=nRT to \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2} } solving to P2 we get:\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}=P_{2} replace the values P_{2}=\frac{2.5*3*610}{9.5*310} =1.55(atm).

5 0
3 years ago
5. Why are ceramics and plastics good insulators?
jeka57 [31]

Answer:

Here

Explanation:

They don't have free electrons moving around (delocalised electrons) so they can't conduct heat and electricity which gives them a property of good insulators. The insulators stop us having an electric shock because they don't conduct electricity as we use them to insulate metal wires and other metallic things. can i have brainliest now pls!

3 0
3 years ago
QUESTION 1<br> 67.2 kilometers = how many decameters
den301095 [7]

Answer: the answer should be 6,720 decameters.

6 0
4 years ago
Read 2 more answers
How much energy is released when the correct number of protons and neutrons come together to form deuterium? a) 0.52 MeV b) 1.7
irga5000 [103]

<u>Answer:</u> The correct answer is 2.24 MeV.

<u>Explanation:</u>

The chemical reaction for the formation of deuterium from proton and neutron follows:

_{1}^1\textrm{H}+_0^1\textrm{n}\rightarrow _1^2\textrm{H}

We are given:

Mass of _{1}^{1}\textrm{H} = 1.00784 u

Mass of _{0}^{1}\textrm{n} = 1.008665 u

Mass of _{1}^{2}\textrm{H} = 2.014102 u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

\Delta m=(m_{_1^2H}+m_{n})-(m_{_1^2H})\\\\\Delta m=(1.00784+1.008665)-(2.014102)=0.002403u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.002403u)\times c^2

E=(0.002403u)\times (931.5MeV)    (Conversion factor:  1u=931.5MeV/c^2  )

E=2.24MeV

Hence, the energy released in the given nuclear reaction is 2.24 MeV.

5 0
3 years ago
An inventor is applying for a patent. He claims his new heat engine can produce 1,200 J of work for every 1,800 J of heat applie
Marat540 [252]

The efficiency of the heat engine is 66.67%.

<h3>Efficiency of the heat engine</h3>

The efficiency of the heat energy is determined by taking the ratio of the output energy to input energy as shown below;

\eta = \frac{0utput \ energy}{1nput \ energy} = \frac{1200}{1800} \times 100\% = 66.67\%

<h3 /><h3>Evaluation of the claim</h3>

The efficiency of the heat engine is greater than 50% but less than 75% and can be considered to be moderately efficient.

This implies that the heat engine can convert up-to two-third of the input energy into useful energy.

Learn more about efficiency here: brainly.com/question/15418098

5 0
3 years ago
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