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3 years ago

Evaluate tan45/tan45

Physics

2 answers:

Andru [333]3 years ago

7 0

Answer:

Explanation:

tan 45° = √2\2

tan θ = y\x

According to the Unit Circle, when you divide √2\2 by

itself, you get 1.

SIZIF [17.4K]3 years ago

5 0

I'm not sure I completely understand the expression you want evaluated.

It looks like a fraction with the same exact thing in both the numerator and the denominator. A fraction like that always boils down to ' 1 '.

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Will give correct answer brainliest 5 kg m/s 8kg m/s 80 kg m/s 200 kg m/s

o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

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3 years ago

A bar of gold has a temperature of 31°C, and a bar of aluminum has a

Katarina [22]

Different dense matters

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3 years ago

A particle cannot generally be localized to distances much smaller than its de Broglie wavelength. This fact can be taken to mea

Vlada [557]

The De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

<h3>How is the De broglie wavelength of a thermal neutron at room temperature calculated?</h3>

Temperature, T = 300K

Momentum, p = mv

Therefore v = p/m

Energy, E= 1/2 m( p/m) ²

Boltzman Energy= 3/2 KT

3/2KT = 1/2 m(p/m)²

Therefore p = $\sqrt{3RTm}$

According to De broglie hypothesis, P = h ÷ λ

Therefore, λ = h ÷ $\sqrt{3RTm}$

= 6.6× $10^{-34}$ ÷ $\sqrt{3 \times 1.38 \times 10^{-23} \times 300 \times 1.6 \times 10^{-27} }$

= 0.15 × $10^{-9}$

Therefore the De broglie wavelength of a thermal neutron at room temperature 300K = 1.5 × A°

To learn more about De broglie wavelength, refer: https://brainly.in/question/6131028

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2 years ago

What does the principal quantum number determine?

morpeh [17]

principal quantum number (n) represents the relative overall energy of each orbital

Hope this helps!

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3 years ago

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ipn [44]

Answer:

answer D

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