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mina [271]
2 years ago
8

you should begin viewing a bacteria specimen with what objective lens? view available hint(s)for part g you should begin viewing

a bacteria specimen with what objective lens? 100x 10x 40x
Physics
1 answer:
Sergio039 [100]2 years ago
5 0
  • Some people view bacteria specimens with a 100x objective lens in order to see the smallest details.
  • Others may use a 10x objective lens for more general purposes, such as examining stained slides or pictures.
  • And still others may use a 40x objective lens to gain maximum resolution when viewing images of thick samples.

It is important to choose the appropriate magnification for your needs so that you can properly examine the specimen under study.

<h3>Why is the 100x objective lens necessary to see bacteria?</h3>
  • Bacteria must, of course, be viewed at the maximum magnification and resolution possible because to their small size.
  • Due to optical restrictions, this is approximately 1000x in a light microscope.
  • To improve resolution, the oil immersion method is performed. This calls for a unique 100x objective.

To learn more about bacterial specimen, visit:

brainly.com/question/1412064

#SPJ4

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Kinetic energy is movement, thus movement would be a characteristic that allows an object to have kinetic energy
6 0
3 years ago
A 41.0 g marble moving at 2.30 m/s strikes a 25.0 g marble at rest. What is the speed of each marble immediately after the colli
Gre4nikov [31]

Answer:

speed of each marble after collision will be 1.728 m/sec

Explanation:

We have given mass of the marble m_1=41gram=0.041kg

Velocity of marble v_1=2.30m/sec

Its collides with other marble of mass 25 gram

So mass of other marble m_2=25gram=0.025kg

Second marble is at so v_2=0m/sec

We have to find the velocity of second marble

From momentum conservation we know that

m_1v_1+m_2v_2=(m_!+m_2)v, here v is common velocity of both marble after collision

So 0.041\times 2.30+0.025\times 0=(0.041+0.025)v

v = 1.428 m /sec

So speed of each marble after collision will be 1.728 m/sec

6 0
3 years ago
Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.
fredd [130]

Energy to lift something =

               (mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm  =  5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

            =  2,696,925 joules .

That's quite a large amount of energy ... equivalent to
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7 0
3 years ago
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I would tell him, in the kindest, most gentle way I could manage,
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You can never get more energy out of the electromotor than you put into it,
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Pour yourself a cold glass of soda, then look up "Perpetual Motion" or
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