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3 years ago

Which is not an optical instrument?

Physics

2 answers:

exis [7]3 years ago

7 0

Answer:

Option (D), television

Explanation:

The optical instrument is an instrument whose working is explained by the phenomenon of reflection or refraction.

An optical instrument consists of mirrors, lenses or both.

Camera, microscope and telescope all are based on the phenomenon of reflection, refraction or both. They consists of lenses or mirrors.

Television is the device in which an electron gun is used. An electric and magnetic field is also used which gives the straight line path to the electron beam.

exis [7]3 years ago

5 0

Out of all the options, Television is not an optical instrument.

In short, Your Answer would be Option D

Hope this helps!

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3 years ago

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3 years ago

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3 years ago

A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

elena-14-01-66 [18.8K]

Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

3 0

3 years ago