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Olenka [21]
3 years ago
15

Helppppp pleaseee :(

Physics
1 answer:
Elza [17]3 years ago
7 0

1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum

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what is the thickness of the central portion of a thin conveying lens can be determined very accurately by using (a) vernier cal
beks73 [17]

Option B The thickness of the central portion of a thin conveying lens can be determined very accurately by using a micrometer screw gauge.

<h3>What can be measured using a micrometer screw gauge?</h3>

One micrometer of thickness can be measured with a micron micrometre screw gauge. A Use of Micrometer Screw Gauge as like example Upon turning the screw of the micrometer screw gauge four times, a 2 mm space is covered.

<h3>What purposes does a micrometer serve?</h3>

A tool known as a micrometer is used to measure solid objects’ lengths, thicknesses, and other dimensions precisely and linearly.

<h3>What is the micrometer screw gauge’s SI unit?</h3>

The SI symbol m is also known as a micron, which is an SI-derived unit of length equaling 1106 meters, where 106 is the SI standard prefix for the prefix “micro-.” A micrometer is one-millionth of a meter.

To know more about screw gauges, visit:

brainly.com/question/4704005

#SPJ13

8 0
11 months ago
What do neon,oxygen and nitrogen have in common?
julsineya [31]
Their Period number is common means their "Principal Quantum Numbers" are same

Hope this helps!
7 0
3 years ago
The specific heat of water is 4,186 J/kg.'C. Approximately how much heat must
Artyom0805 [142]

Explanation:

Q= mc∆T

∆T= 5-24=- 19

Q= 0.5*4186*-19

Q= -39767 J

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4 0
2 years ago
Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The
Rudik [331]

Answer:

1)   P₁ = -2 D,   2) P₂ = 6 D

Explanation:

for this exercise in geometric optics let's use the equation of the constructor

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where f is the focal length, p and q are the distance to the object and the image, respectively

1) to see a distant object it must be at infinity (p = ∞)

          \frac{1}{f_1} = \frac{1}{q}

           q = f₁

2) for an object located at p = 25 cm

            \frac{1}{f_2} = \frac{1}{25} + \frac{1}{q}

We can that in the two expressions we have the distance to the image, this is the distance where it can be seen clearly in general for a normal person is q = 50 cm

we substitute in the equations

1) f₁ = -50 cm

2)  

        \frac{1}{f_2} = \frac{1}{25} + \frac{1}{50}

        \frac{1}{f_2} = 0.06

         f₂ = 16.67 cm

the expression for the power of the lenses is

          P = \frac{1}{f}

where the focal length is in meters

           

1)       P₁ = 1/0.50

        P₁ = -2 D

2)     P₂ = 1 /0.16667

        P₂ = 6 D

4 0
2 years ago
Write the chemical formula for the following diagrams.
seraphim [82]

Answer:

hydrogen chloride...........

4 0
3 years ago
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