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ohaa [14]
3 years ago
6

HELP PLEASE! I have some homework for my 9th grade physical science class and it's due Monday, I already completed 1/2 of it, if

you can answer one or more that'll be really helpful, ty! If a question does not make sense please let me know, each of the pages contain a small paragraph that may be needed to complete the question. (p.s. if u do answer a question, please identify the page it is on and what number it is, tysm)
page 1
3: The symbol for one of the elements is derived from its German name, wolfram. What is the English name for this element?

page 2
1. What limitation is placed on electrons in the Bohr model of the atom?

2. When the light from a gas is passed through a prism, why is there only a certain number of lines rather than a continuous spectrum?

3. How is the series of lines emitted by a gas similar to a fingerprint?

4. Explain how scientists can use the pattern of lines from an unknown gas to identify what elements are contained in the gas?

page 3
1. what is ONE important difference between the Greek theory of atoms and Dalton's atomic theory?

2. what is the significance of the invention of the STM?

3. explain how an STM creates an image of the atoms on the surface of a material
Physics
1 answer:
faltersainse [42]3 years ago
8 0

Page 1: #3

Wolfram is known as Tungsten in English (Tungsten's symbol is a W).

Page 2: #1

I think the limitation is that the electrons in the Bohr model are only in certain energy levels and that they orbit the nucleus.

It's been a while since I was in a physical science class so I'm not sure about the rest but I wish you luck! I hope this helps a little bit!

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b) 19.82 m/s

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t = Time taken

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⇒Distance = (75/3.6)×0.6

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v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-\frac{75}{3.6}^2}{2\times -8.5}\\\Rightarrow s=25.53\ m

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v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-u^2}{2\times -8.5}\\\Rightarrow s=\frac{u^2}{17}

d + s = 35

\\\Rightarrow u0.6+\frac{u^2}{17}=35\\\Rightarrow \frac{u^2}{17}+0.6u-35=0

10u^2+102u-5950=0\\\Rightarrow u=\frac{-51+\sqrt{62101}}{10},\:u=-\frac{51+\sqrt{62101}}{10}\\\Rightarrow u=19.82, -30.02

Hence, the maximum velocity by which the car could be moving and not hit the barrier 35 meters ahead is 19.82 m/s.

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2 years ago
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