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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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One kilowatt-hour = 1000 watts = 1000 joules/second
One hour = 60 × 60 = 3600 seconds

One kWh = 3600 × 1000 joules
One kWh = 36 × 100000 joules
One kWh = 3.6 × 10 × 100000 joules
One kWh = 3.6 × 10⁶ joules

Answer: Second option
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3 years ago
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What are the advantages of using a telescope compared to making observations with the naked eye?
lara31 [8.8K]
The naked eye is very limited. Unlike the telescope which can see far away.
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Two lab carts are pushed together with a spring mechanism compressed between them. Upon release, the 5.0 kg cart repels one way
JulijaS [17]

Answer:

V = -0.3 m/sec.

Explanation:

5.0 x 0.12 + 2.0 x v = 0. Which means that V = -0.3 m/sec.

The -ve sign shows it moves in the opposite direction.

8 0
3 years ago
In a simple electric circuit, Ohm's law states that V=IRV=IR, where V is the voltage in volts, I is the current in amperes, and
VMariaS [17]

To solve this problem, apply the concepts given from Ohm's Law. From there we will obtain the derivative of the function with respect to time and with the previously given values we will proceed to find the change in current as a function of the derivative

V = IR

Here

I = Current

V = Voltage

R = Resistance

Taking the derivative we will have,

\frac{dV}{dt} = \frac{dI}{dt}R + I \frac{dR}{dt}

Our values are given as,

\frac{dV}{dt} = -0.01

\frac{dR}{dt} = 0.04\Omega/s

R = 300\Omega

I = 0.01A

Replacing we will have that

\frac{dV}{dt} = \frac{dI}{dt}R + I \frac{dR}{dt}

-0.01 = \frac{dI}{dt}(300\Omega) + (0.01A)(0.04\Omega/s)

Rearranging to find the current through the time,

\frac{dI}{dt} = \frac{-0.01- (0.01A)(0.04\Omega/s)}{(400\Omega)}

\frac{dI}{dt} = -0.000026A/s

Therefore the change of the current is -0.000026A per second

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3 years ago
On a sky coaster (human pendulum) that reaches 20 meters from it's equilibrium position, a man of 70 kg is able to reach a maxim
Reika [66]

Answer:

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

V

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 20 m for h then

v=√(2*9.81*20)=19.799 m/s

Rounding off, v is approximately 20 m/s

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