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3 years ago

I need help with 1-7 ASAP

Physics

1 answer:

elena55 [62]3 years ago

5 0

Refraction. ... Diffraction. ... EM spectrum. ... Intensity. ... Transverse wave. ... Frequency. ... Compression wave.

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Do noble gases lose or gain electrons to get a full valence shell?

alexdok [17]

They don't lose OR gain electrons as they've already achieved the octet rule and have 8 valence electronsn

8 0

3 years ago

State a true conclusion.

Ganezh [65]

The conclusion is; Cathy's dad has gone fishing.

<h3>What are conclusions in conditional statements?</h3>

A conditional statement is a statement with a hypothesis which is then followed by a conclusion.

A conditional statement has two parts; "if" and "then"

The hypothesis is “if,” part of a conditional statement.

The conclusion is the “then,” part of a conditional statement.

Considering the given conditional statements:

1. If it's Saturday, then Cathy's dad goes fishing.

2. If it's Saturday, then Cathy's dad goes fishing.

Since the if part is true, the conclusion will be, Cathy's dad has gone fishing.

In conclusion, a conditional statement consists of a hypothesis and a conclusion.

Learn more about conditional statements at: brainly.com/question/21170
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7 0

2 years ago

\Which are consumers within a spruce-fir forest ecosystem?

lara [203]

The answer is mushroom.

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3 years ago

Read 2 more answers

Areas near oceans have ______________________ than areas in the interior of continents because of the great storage capacity of

zhenek [66]

The answer would be more mass

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3 years ago

What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g

aalyn [17]

Answer:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle $\theta$ , given by:

$f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)$

Where:

$Z_1 e$ represent the charge of the projectile (Z1=2)

$Z_2 e$ is the target charge (z2=79)

$K= 5x10^6 eV$ represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

$t= 10^{-8] m$ represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

$n =\frac{\rho N_A N_M}{M_g}$

Where:

$\rho = 19.3 g/m^3= 19300 Kg/m^3$

$N_A = 6.022 x10^{23} molecules/mol$ the Avogadro's number

$N_M = 1$ represent the atoms per molecule

$M_g = 197 g/mol = 0.197 Kg/mol$ represent the molecular weigth

If we replace we got:

$n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3$

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness $t=10^{-8}m$

And using the first formula we got:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.

4 0

3 years ago