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Gemiola [76]
2 years ago
13

A brightfield microscope equipped with a ____________ ___________ stop is able to block light from the illuminator creating brig

ht objects on a dark background.
Physics
1 answer:
Arturiano [62]2 years ago
4 0

Answer: A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.

Explanation: To find the correct statement about Brightfield microscope, we need to know more about the Brightfield microscope.

<h3>what is Brightfield microscope?</h3>
  • The Brightfield microscopic technique is generally used with the compound microscopes and, these are one, among the widely used optical illumination techniques.
  • The Brightfield microscope or Compound light microscope is an optical instrument, which uses light rays and produces dark images with a bright background.
<h3>How Bright field microscope works?</h3>
  • A brightfield microscope equipped with a opaque light stop, will produce bright objects on a dark background by blocking the light illuminated from the light source.
  • This opaque light stop, blocks the light rays that arising from the illuminator to the objective lenses and, only allows the reflected or transmitted light of the specimen. As a result, we get bright objects with a dark background.

Thus, we can conclude that, A Brightfield microscope equipped with an opaque light stop is able to block light from the illuminator and creating bright objects on a dark background.

Learn more about the Brightfield microscope here: brainly.com/question/28019837

#SPJ4

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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
Tresset [83]

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

 So

       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

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Explanation:

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In short, Your Answer would be "Decreases"

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