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Varvara68 [4.7K]
3 years ago
7

Question 1 of 10

Physics
1 answer:
Marianna [84]3 years ago
6 0

Answer:

Option D. ²²²₉₀Th

Explanation:

Let the unknown be ⁿₘZ. Thus, the equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

Next, we shall determine n, m and Z. This can be obtained as follow:

For n:

226 = 4 + n

Collect like terms

226 – 4 = n

222 = n

n = 222

For m:

92 = 2 + m

Collect like terms

92 – 2 = m

90 = m

m = 90

For Z:

ⁿₘZ => ²²²₉₀Z => ²²²₉₀Th

Therefore, the complete equation becomes:

²²⁶₉₂U —> ⁴₂He + ⁿₘZ

²²⁶₉₂U —> ⁴₂He + ²²²₉₀Th

Thus, the unknown is ²²²₉₀Th

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Answer:

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Explanation:

Given that,

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F_{net}=F_{in}-F_{out}

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F_{net}=(1.90\times10^{5}-1.013\times10^{5})3.90\times10^{-4}

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(b) an astronomical unit (au) is the average distance from the sun to earth, 1.50 × 108 km. how many au are there in 11.0 light-years?

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We want to design a cylindrical vacuum capacitor, with a given radius a for the outer cylindrical shell, that will be able to st
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Solution :

a). Using Gauss's law :

  $E=\frac{Q}{4 \pi \epsilon_0r^2}$  ,    $b    .........(1)

Let $E=E_0,\ r=b$ in equation (1)

Therefore, $Q=4 \pi \epsilon_0b^2E_0$  .............(2)

$V_b-V_a = \int^a_b \vec E. d\vec l$

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$b=\frac{3}{4}a$  

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$U = 4 \pi\epsilon_0 \frac{27a^3E^2_0}{512}$

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