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SVEN [57.7K]
3 years ago
11

12) If, after viewing a specimen at low power, you switch to high-dry power and, after using fine focus, cannot find the specime

n, what things could you do to help yourself (before calling me over to assist you?)
Physics
1 answer:
Annette [7]3 years ago
3 0

Answer:

See the answer below

Explanation:

After seeing an object on a slide at the low-power objective of the microscope and it disappears on changing to high power, the following can be done to resolve the problem

1. <em>Drop a few drops of immersion oil on the slide and view again under high the power objective.</em>

2. <em>If the object is still not visible after the action above, return the microscope to the low-power objective and make sure the object is refocused and centered. Then carefully change back to the high power objective and use the fine adjustment to bring it into focus.</em>

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What happens when white light shines onto a red filter?
Gnesinka [82]
The dye in between the two pieces of glass in the filter absorbs all the other colors in the white light except the red, and the red is the only light left to come out the other side. That's why the filter usually looks kind of red. It's also a big part of the reason why we call it a "red filter".
6 0
3 years ago
Particles q1, 92, and q3 are in a straight line.
NNADVOKAT [17]

The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃

<h3 /><h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The force,by the charge q₁ on the q₃;

\rm F_{31}} = \frac{Kq_1q_3}{r^2} \\\\ \rm F_{31}}  = \frac{9 \times 10^9 \times -28.1 \times 10^{-6}\times \times -47.9 \times 10^{-6}}{(0.600)^2} \\\\ F_{31}} =33.64 \ N

The force,by the charge q₂ on the q₃;

\rm F_{32}} = \frac{Kq_2q_3}{r^2} \\\\ \rm F_{32}}  = \frac{9 \times 10^9 \times 25.5 \times 10^{-6}\times \times -74.9\times 10^{-6}}{(0.300)^2} \\\\ F_{32}} =-19.09  \ N

The net force is the sum of the two forces;

\rm F_{net}=F_{32}+F_{31}\\\\\ \rm F_{net}=36.6-19.9 \\\\ \rm F_{net}=17.51 \ N

Hence, the net force on q₃ will be 17.51 N.

To learn more about Columb's law, refer to the link;

brainly.com/question/1616890

#SPJ1

8 0
2 years ago
the space shuttle used three parachutes to land in the image at the start of the project. Why did the space shuttle engineers us
iragen [17]

Answer:

Sep 9, 2008 - Image above: In the Parachute Refurbishment Facility at NASA's ... His parachute, when opened, had a canopy that resembled a huge umbrella. ... Space Center used mainly to process space shuttle parachutes, United Space ... Hillebrandt also is lead engineer for the Constellation Program's Ares 1 rocket ...

Explanation:

Sep 9, 2008 - Image above: In the Parachute Refurbishment Facility at NASA's ... His parachute, when opened, had a canopy that resembled a huge umbrella. ... Space Center used mainly to process space shuttle parachutes, United Space ... Hillebrandt also is lead engineer for the Constellation Program's Ares 1 rocket ...

6 0
3 years ago
A beam of visible light is passed through a plexiglass container filled with water. If a beam of light comes in at a slight angl
pentagon [3]

B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
8 0
3 years ago
Read 2 more answers
a sprinter running a 100m dash leaves the starting block and accelerates to a maximum velocity of 11m/s at 6s into the race. the
mihalych1998 [28]

Answer:

Part(a): The average acceleration is 1.83~m~s^{-2} during the first 6s.

Part(b): The average acceleration from 6s to 8s is zero.

Explanation:

Part(a):

The average acceleration is defined as the rate at which velocity is changing with respect to time. So if 'v_{i}', 'v_{f}' and 'a_{av}' represents the initial velocity, final velocity, and average acceleration of a particle, then mathematically

a_{av} = \dfrac{v_{f} - v_{i}}{t}

where 't' is the time taken by the particle to achieve the velocity 'v_{f}' starting from initial velocity 'v_{i}'

Given in the problem, v_{i} = 0, v_{f} = 11~m~s^{-1}~and~t = 6~s.

So the average acceleration(a_{av}) during the first 6 s will be

a_{av} = \dfrac{11 - 0}{6}~m~s^{-2} = 1.83~m~s^{2}

Part(b):

During the time between 6 s to 8 s, as mentioned in the problem, the sprinter maintains the constant velocity. So the average acceleration during this time interval will be zero.

7 0
3 years ago
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