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2 years ago

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

Physics

1 answer:

Natalka [10]2 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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A photographer in a helicopter ascending vertically at a constant rate of accidentally drops a camera out the window when the he

maksim [4K]

Answer:

The time it takes for the camera to reach the ground is 5 s.

Explanation:

To solve this problem, we will use the free fall cinematic equation.

Since the helicopter ascends with constant speed, the camera falls to the ground only by the effect of gravity on it.

The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 $\frac{m}{s}$ .

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

To calculate the maximum height to which it arrives, we will use the formula of vertical shot (since the camera leaves the helicopter with a speed upwards of 12,5 $\frac{m}{s}$ ).

$V_{f} ^{2}$ = $V_{0} ^{2}$ - 2 * g * h

Where:

$V_{f}$ : final speed at maximum height.

$V_{0}$ : initial speed when it falls from the helicopter.

g: gravity taken at 9.8 $\frac{m}{s^{2} }$

h: height reached from 60 m when leaving the helicopter

as $V_{f}$ =0

0= $(12,5\frac{m}{s}) ^{2}$ - 2 * $9,8\frac{m}{s^{2} }$ * h

clear h:

h= $(12,5 \frac{m}{s^{2} } )^{2}$ / (2 * $9,8 \frac{m}{s^{2} }$ )

h=7,97 m

Then we must calculate the time it takes to reach its maximum height:

$V_{f}$ = $V_{0}$ - g * t

t: time it takes to arrive from the moment it leaves the helicopter at its maximum height.

as $V_{f}$ =0

0=12.5 $\frac{m}{s}$ - 9.8 $\frac{m}{s^{2} }$ * t

clearing t

t=12.5 $\frac{m}{s}$ / 9.8 $\frac{m}{s^{2} }$

t=1.27 s.

Now we can calculate the time it takes to fall from the maximum height of 67.97 m.

The equation we will use is Y= $v_{0}*t+(\frac{g*t^{2} }{2} )$

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

$v_{0}$ : initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 $\frac{m}{s^{2} }$

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

$v_{0}$ =0

So the first term of our equation is nullified.

Step 2: To calculate the time it takes to fall, we clear "t" of the equation:

Y= $\frac{(g*t^{2})}{2}$

Y*2=(g* $t^{2}$ )

$\frac{Y*2}{g}$ = $t^{2}$

$\sqrt{\frac{Y*2}{g} }$ =t

Step 3: I replace the values with the incognites and get "t".

t= $\sqrt{\frac{67,97m*2}{9,8\frac{m}{s^{2} } } }$

t=3,73 s

The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!

3 0

2 years ago