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VladimirAG [237]
1 year ago
13

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

liper (b) micro meter screw guage (c) telescope ( d) microscope​
Physics
1 answer:
beks73 [17]1 year ago
8 0

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

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Compare the time period of two simple pendulums of length 4m and 16m at a place.
Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0
3 years ago
a vector is 253 m long and points in a 55.8 degree direction, what’s the y and x- component of the vector?
BartSMP [9]

Well we know the hypotenuse of the triangle which is 253 m. And we know the angle of the triangle which is 55.8 degrees. So we want to find y. And to find y we use sin. And sin is a ratio, the ratio of the opposite leg, and hypotenuse. So sin(55.8) = y/253. Now we solve for y by multiplying both sides by 253. And finally we get 209.25 as the length of the y component.

5 0
3 years ago
What is the displacement of the student from the bus stop to the mailbox?
Reika [66]

Answer:C

Explanation:

5 0
3 years ago
An airplane has an effective wing surface area of 19.4 m2 that is generating the lift force. In level flight the air speed over
alexgriva [62]

Answer:

W =23807.68 N

Explanation:

given,

surface area of wing = 19.4 m²

speed over top wing = 67 m/s

speed under wing = 51 m/s

density of air =  1.3 kg/m³

weight of plane

From Bernoulli's principle

P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho_2^2

where 1 and 2 are two different locations at the same geo potential level  

so if we call 1 the lower surface and 2 the upper surface,

we find the pressure differential, P₁ -P₂

\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)

\Delta P =\dfrac{1}{2}\times 1.3 \times (67^2-51^2)

\Delta P =1227.2\ N/m^2

then the force acting on the plane is

F=P A

F=1227.2 x 19.4

F =23807.68 N

weight of the plane

W =23807.68 N

7 0
3 years ago
It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³
Evgesh-ka [11]

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, \rho_o = 800 kg/m³

density of water, \rho_w = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

8 0
3 years ago
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