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IgorLugansk [536]
3 years ago
8

Can anyone tell me how to read a micrometer screw gauge I want very clear instructions.

Physics
1 answer:
Natalka [10]3 years ago
5 0

Explanation:

Things you need to know:

Accuracy refers to the maximum error encountered when a particular observation is made.

Error in measurement is normally one-half the magnitude of the smallest scale reading.

Because one has to align one end of the rule or device to the starting point of the measurement, the appropriate error is thus twice that of the smallest scale reading.

Error is usually expressed in at most 1 or 2 significant figures.

Tape

Equipment: It is made up of a long flexible tape and can measure objects or places up to 10 – 50 m in length. It has markings similar to that of the rigid rule. The smallest marking could be as small as 0.1 cm or could be as large as 0.5 cm or even 1 cm.

How to use: The zero-mark of the measuring tape is first aligned flat to one end of the object and the tape is stretched taut to the other end, the reading is taken where the other end of the object meets the tape.

Ruler

Equipment: It is made up of a long rigid piece of wood or steel and can measure objects up to 100 cm in length. The smallest marking is usually 0.1 cm.

How to use: The zero-end of the rule is first aligned flat with one end of the object and the reading is taken where the other end of the object meets the rule.

Vernier Caliper

Equipment: It is made up of a main scale and a vernier scale and can usually measure objects up to 15 cm in length. The smallest marking is usually 0.1 cm on the main scale.

It has:

a pair of external jaws to measure external diameters

a pair of internal jaws to measure internal diameters

a long rod to measure depths

How to use: The jaws are first closed to find any zero errors. The jaws are then opened to fit the object firmly and the reading is then taken.

Micrometer Screw Gauge

Equipment: It is made up of a main scale and a thimble scale and can measure objects up to 5 cm in length. The smallest marking is usually 1 mm on the main scale (sleeve) and 0.01 mm on the thimble scale (thimble). The thimble has a total of 50 markings representing 0.50 mm.

It has:

an anvil and a spindle to hold the object

a ratchet on the thimble for accurate tightening (prevent over-tightening)

How to use: The spindle is first closed on the anvil to find any zero errors ( use the ratchet for careful tightening). The spindle is then opened to fit the object firmly (use the ratchet for careful tightening) and the reading is then taken.

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pogonyaev
Displacement is usually given to you as it is, but you can also get displacement through velocity by Δd= Δv*t, where  <span>Δv is the change in velocity and t is the change in time. 

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4 0
3 years ago
An archer pulls her bowstring back 0.396 m by exerting a force that increases uniformly from zero to 237 N. (a) What is the equi
tatyana61 [14]

Answer:

(a) The equivalent spring constant is 598.485 N/m

(b) The work done is 46.926 J

Explanation:

From Hooke's law of elasticity

K (spring constant) = F/e

F is the range of force exerted = 237 - 0 = 237 N

e is the extension of bowstring = 0.396 m

K = F/e = 237/0.396 = 598.485 N/m

Work done = 1/2 Fe = 1/2 × 237 × 0.396 = 46.926 J

6 0
3 years ago
The specific heat capacity of solid copper metal is 0.385 J/gK. How many joules of heat are needed to raise the temperature of a
SSSSS [86.1K]

Answer: 26.6 J

Explanation:

The heat needed to raise the temperature of a solid body, using only  a conductive process, has been empirically showed to be equal to the following expression:

Q= c . m.  (t2 – t1)

where c= specific heat capacity (in J/gK), m= mass of the solid (in g) ,

and (t2 - t1)= difference between final and initial temperatures.

Replacing by the values, we get:

Q= 0.385 J/gK . 1,550 g. (77.5ºC – 33.0ºC)= 26.6 J

3 0
3 years ago
How could you increase power of a wave in a spring
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Alll of the above I just read it from someone and if it’s not that lmk
8 0
3 years ago
Read 2 more answers
How much the lengths of various substances change with temperature is given by their coefficients of linear expansion, α. The gr
gtnhenbr [62]

Answer:

aluminium > copper > steel            (amount which get sag)

Explanation:

<u>Theory</u>

Linear expansivity (α) : The increase in length, per unit length per degree rise in temperature.

Therefore, α = Δl/Lθ

                   Δl = increase in length ( amount of sag in this case )

                    L = length of wire

                    θ = temperature change

We get,    Δl = Lαθ

From that we get,    amount of sag ∝ Linear expansivity (α)

Initial length of all three wires are the same.

The temperature change they subject also the same.

So the factor that changes the amount of sag is the coefficient of linear expansivity.  

6 0
2 years ago
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