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3 years ago

Nonmetals are usually , which means that they evaporate easily.

Physics

2 answers:

IrinaVladis [17]3 years ago

8 0

The answer is 'volatile'

Jlenok [28]3 years ago

6 0

Answer:

what he said

Explanation:

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Which way do the players on the court rotate during a game? What happens if the players serve out of order?

tigry1 [53]

Answer:

Explanation:

on volleyballit's clockwise and What Would Happen If Someone Served Out Of Order? If the wrong player on your team serves because it's not their turn, the point and the possession of the serve will go to your opponents.

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3 years ago

PLEASE HELP ME

hammer [34]

Answer:

Element Is The Answer I think

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3 years ago

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The blackbody radation emmitted from a furnace peaks at a wavelength of 1.9 x 10^-6 m (0.0000019 m). what is the temperature ins

krek1111 [17]

Answer:

Temperature, T = 1542.10 K

Explanation:

It is given that,

The black body radiation emitted from a furnace peaks at a wavelength of, $\lambda=1.9\times 10^{-6}\ m$

We need to find the temperature inside the furnace. The relationship between the temperature and the wavelength is given by Wein's law i.e.

$\lambda\propto \dfrac{1}{T}$

$\lambda=\dfrac{b}{T}$

b = Wein's displacement constant

$\lambda=\dfrac{2.93\times 10^{-3}}{T}$

$T=\dfrac{2.93\times 10^{-3}}{\lambda}$

$T=\dfrac{2.93\times 10^{-3}}{1.9\times 10^{-6}\ m}$

T = 1542.10 K

So, the temperature inside the furnace is 1542.10 K. Hence, this is the required solution.

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4 years ago

Which of the following is an ecosystem

iogann1982 [59]

A it is Green plants.

or they all

7 0

3 years ago

Helium–neon laser light (λ = 632.8 nm) is sent through a 0.350-mm-wide single slit. What is the width of the central maximum on

Cloud [144]

Answer:

The width of the central bright fringe is 7.24 mm.

Explanation:

Given that,

Wavelength = 632.8 nm

Width d= 0.350 mm

Distance between screen and slit D= 2.00 m

We need to calculate the distance

Using formula of distance

$y_{m}=\dfrac{\lambda D}{d}$

Put the value into the formula

$y_{m}=\dfrac{632.8\times10^{-9}\times2.00}{0.350\times10^{-3}}$

$y_{m}=3.62\ mm$

We need to calculate the width of the central bright fringe

Using formula of width

$width = 2\times|y_{m}|$

Put the value into the formula

$width=2\times3.62$

$width = 7.24\ mm$

Hence, The width of the central bright fringe is 7.24 mm.

5 0

4 years ago