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yanalaym [24]
3 years ago
13

Examples of black body radiators? Please

Physics
1 answer:
Anton [14]3 years ago
8 0
A cavity that has a hole in it......it's the best example of black body radiators....
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Can anyone tell me how to read a micrometer screw gauge I want very clear instructions.
Natalka [10]

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.

5 0
3 years ago
Goo.gle meet??<br><br> https://meet.go ogle.com/qzz-pxqd-gaj
Elenna [48]

Answer:

ok............I'll try ............

5 0
3 years ago
Philosophers formulated explanations about the nature of matter based on<br> their own experiences.
Galina-37 [17]

Answer:

Both Democritus and Dalton suggested that matter is made up of atoms. ... Dalton's atomic theory stated that atoms of different elements combine in simple whole-number ratios to form compounds.

4 0
3 years ago
The conductive tissues of the upper leg can be modeled as a 40-cm-long, 12-cm-diameter cylinder of muscle and fat. The resistivi
ahrayia [7]

To calculate and solve the problem it is necessary to apply the concepts related to resistance and resistivity.

The equation that is responsible for relating the two variables is:

R = \rho \frac{L}{A}

Where,

R= Resistance of the conductor

\rho =Resistivity of the conductor material

L = Length

A = Cross-sectional area of conductor

With the previous values the area of the muscle (Real Muscle-82%)is,

A_m = (0.82)\pi r^2 = (0.82)\pi (12/2*10^{-2})^2

A_m = 9.274*10^{-3}m^2

Using the equation from Resistance we have that at the muscle the value is:

R_m = \rho \frac{L}{A}

R_m = \frac{13*(0.4)}{9.274*10^{-3}}

R_m = 560.70\Omega

At the same time we can make the same process to calculate the resistance of the fat, then

A_m = (0.18)\pi r^2 = (0.18)\pi (12/2*10^{-2})^2

A_m = 2.0357*10^{-3}m^2

The resistance of the fat would be,

R_f = \rho \frac{L}{A}

R_f = \frac{25*(0.4)}{2.0357*10^{-3}}

R_f = 4912.3\Omega

Then the total resistance in a set as the previously writen, i.e, in parallel is:

R=\frac{R_mR_f}{R_m+R_f}

R = \frac{(560.70)(4912.3)}{4912.3+560.70}

R = 502.62\Omega

We can here apply Ohm's law, then

I= \frac{V}{R}

I = \frac{1.5}{502.62}

I = 2.984*10^{-3}A

I = 2.984mA

3 0
4 years ago
A 2.0kg rocket is traveling straight up. What force muse be applied to the rocket for it to move at a constant velocity?
Yuliya22 [10]
A equal force it always has to be equal
5 0
4 years ago
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