All categories

3 years ago

What are three differences and similarities between a Vernier Calipers and a Micrometer? (please leave a good answer)

Physics

2 answers:

Fiesta28 [93]3 years ago

8 0

Answer:

Explanation:

Vernier scale

Mathematical tool

A vernier scale, named after Pierre Vernier, is a visual aid to take an accurate measurement reading between two graduation markings on a linear scale by using mechanical interpolation; thereby increasing resolution and reducing measurement uncertainty by using vernier acuity to reduce human estimation error. Wikipedia

Purpose: Measuring more precisely than could be done unaided when reading a uniformly divided straight or circular measurement scale

Creator: Pierre Vernier

Year invented: 1631

..............................................................................................................................................................................................................................................................................................

A micrometer is a measuring instrument that can make extraordinarily precise measurements. Most micrometers are designed to measure within one one-thousandth of an inch! That's a close fit. Exact measurements like this are necessary when even the smallest of space between objects can cause problems or difficulties.

Micrometer

Mathematical tool

A micrometer, sometimes known as a micrometer screw gauge, is a device incorporating a calibrated screw widely used for accurate measurement of components in mechanical engineering and machining as well as most mechanical trades, along with other metrological instruments such as dial, vernier, and digital calipers. Wikipedia

Operation supported: Component measurement

Purpose: Precision measurement of a component

Variant: Digital mics, Bench micrometer, Limit mics, Digit mics, MORE

Component: Screw, Spindle, Ratchet stop, Anvil, Frame, Barrel, Thimble lock, Thimble

Application: Mechanical Engineering, Machining

anastassius [24]3 years ago

3 0

Answer:

differences:

The resolution. A micrometer provides smaller resolution compared to calipers. A micrometer can measure down to 0.001mm (micron size), while a caliper can measure down to 0.02 only in general. Resolution is the smallest increment that can be measured.

Range. A micrometer has different ranges. The most used range is 1 inch. However, there are available in 2inch, 3inch, 4inch, 5inch, 6inch, etc. While a caliper has a fairly uniform size.

How They Work. The way mechanical works is by the help of the differential screw mechanism, while the vernier caliper uses vernier scale. The digital micrometers also work differently from the digital calipers.

similarities:

Both are simple to use and easy for measurement.

Both Vernier caliper and micrometer screw gauge might have zero error to some extent. However, the zero error can be calculated and removed.

Both have ability to measure small dimensions

Explanation:

Hope it helps and have a great day/night! ^w^

You might be interested in

El tren Lima la Orolla va a una velocidad de 10 km/h y de pronto aplica el freno por un derrumbe en la via. Si demora 18 segundo

Naya [18.7K]

Answer:

Distancia = 50.04 metros

Explanation:

Dados los siguientes datos;

Velocidad = 10 km/h

Tiempo = 18 segundos

Para encontrar la distancia;

Conversión:

10 km/h = 10 * 1000/3600 = 2.78 m/s

Distancia = velocidad * tiempo

Distancia = 2.78 * 18

Distancia = 50.04 metros

Por lo tanto, el tren viajaría 50.04 metros antes de detenerse por completo.

7 0

3 years ago

A +3.4 x 10-6 C test charge experiences forces from two other nearby charges: a 3 N force due east and a 15 N force due west. Wh

Alecsey [184]

Answer:

3.53×10⁶ N/c due west

Explanation:

From the question

E = F'/q........................ Equation 1

Where E = Electric Field, F = Net Force, q = Charge.

But,

F' = F₂-F₁...................... Equation 2

Substitute equation 2 into equation 1

E = (F₂-F₁)/q................ Equation 3

Given: F₁ = 3 N due east, F₂ = 15 N due west, q = 3.4×10⁻⁶ C

Substitute these values into equation 1

E = (15-3)/(3.4×10⁻⁶)

E = 12/(3.4×10⁻⁶)

E = 3.53×10⁶ N/c due west

4 0

3 years ago

The mass of an object can change if

Serhud [2]

D. none of the above

8 0

3 years ago

Read 2 more answers

A marble runs off the edge of a table that is 1.5 m high and the marble lands 0.50 m from the base of the table. a. How much tim

Free_Kalibri [48]

Answer:

t = 0.55[sg]; v = 0.9[m/s]

Explanation:

In order to solve this problem we must establish the initial conditions with which we can work.

y = initial elevation = - 1.5 [m]

x = landing distance = 0.5 [m]

We set "y" with a negative value, as this height is below the table level.

in the following equation (vy)o is equal to zero because there is no velocity in the y component.

therefore:

$y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]$

Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.

$(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]$

3 0

4 years ago

the speed of a car decreases uniformly from 30 m/s to 10 m/s in 4.0 seconds. the car's acceleration is-

VMariaS [17]

U=30m/s
v=10m/s
t=4s
a=10-30/4
a=-5 m/second square

4 0

3 years ago