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Sidana [21]
1 year ago
5

When using higher magnification, you need more light. which parts of the microscope can be adjusted to provide more light, and h

ow would you adjust those parts to increase the light?
Physics
1 answer:
valina [46]1 year ago
8 0

Increased lamp voltage is achieved by turning the light intensity dial.

To enlarge the diameter of the hole and let more light through the slide, the iris diaphragm was modified.

Condenser: Position it higher and closer to the slide's bottom to better direct light to the centre of the slide.

<h3>How do you adjust the light level on a microscope?</h3>

Utilize the brightness adjustment knob to change the brightness. Turn the brightness control knob while looking through the eyepieces to make sure there is no glare in the field of view.

Use a daylight balancing filter if your compound microscope has a certain sort of illumination. It typically rests directly on top of the luminator or in a filter holder above the light. This filter is blue.

The daylight balancing filter will correct the colour temperature and produce a higher-quality image if your microscope is lighted by tungsten or halogen (and a better colour image). This blue filter is not necessary if your microscope is an LED.

To learn more about light level on a microscope, visit:

brainly.com/question/14727797

#SPJ4

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Which of the following is an idea that robert hutchings goddard introduced in his “further developments” to his research "a meth
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3 years ago
the average speed of a runner in a 483. meter race is 3.0 meters per second. How long me runner to complete the race? Dont inclu
lara [203]

Answer:

161

Explanation:

v=\frac{d}{t} slove for t

t=\frac{d}{v}

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t=\frac{483}{3} \\

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3 0
2 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
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