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kakasveta [241]
3 years ago
7

It would be really helpful if u help me solving this question. PLEASE!!!

Physics
1 answer:
sweet [91]3 years ago
7 0

Answer: The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

The room temperature will be the point at which the themometric substance remains constant when rising from ice point.

Explanation:

Apparatus available:

Unmarked thermometer

250 cm3 glass beaker

crushed ice 

water

heatproof mat 

clamp, boss and stand

meter rule

Added apparatus

Bunsen burner

Stirrer

Method

The students will determine the two fixed points of the thermometer:

Lower fixed point = 0 degree Celsius

Upper fixed point = 100 degree Celsius

Then divide the thermometer with equal intervals

Procedures

Set up the apparatus of illustrated in the attached figure.

Immerse the unmarked thermometer into the ice in the beaker.

When the level indicated by the thermometric substance remains steady after some time, a mark will be made at that point. This mark will corresponds to the ice point (lower fixed point) and is assigned the value of 0 °C.

You may add little water and continue to stir gently.

The themometric substance will start to rise and stop when it reaches room temperature. Mark the point but do not assign any value

Place the beaker on bunsen burner and boil the water. The themometric substance will continue to rise and remain constant at upper fixed point

This mark will corresponds to the steam point (upper fixed point) and is assigned the value of 100 °C.

Divide between the lower fixed point and upper fixed point into equal intervals. Then you can see the value of room temperature.

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Answer:

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A polarizer changes the orientation of the oscillations of a light wave.

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Intensity of light

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⇒I = 10cos²60

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3 years ago
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A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

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Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

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Using formula of tunnel barrier

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Put the value into the formula

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(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

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Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

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