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seropon [69]
3 years ago
13

In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os

cillations. What is the maximum current through the inductor during the oscillations?
A) 3.7 A
B) 42 A
C) 2.5 A
D) 10 A
E) 6.5 A
Physics
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

\omega=\dfrac{1}{\sqrt{CL}}

By putting the values

\omega=\dfrac{1}{\sqrt{CL}}

\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}

ω  = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33  A

I= 6.49 A

I = 6.5 A

E) 6.5 A

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Polonium, the Period 6 member of Group 6A(16), is a rare radioactive metal that is the only element with a crystal structure bas
Burka [1]

Answer:

Approximate atomic radius for polonium-209 is 167.5 pm .

Explanation:

Number of atom in simple cubic unit cell = Z = 1

Density of platinum = 9.232 g/cm^3

Edge length of cubic unit cell= a = ?

Atomic mass of Po (M) = 209 g/mol

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

ρ = density

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

On substituting all the given values , we will get the value of 'a'.

9.232 g/cm3=\frac{1 \times 209 g/mol}{6.022\times 10^{23} mol^{-1}\times (a)^{3}}

a = 3.35\times 10^{-8} cm

Atomic radius of the polonium in unit cell = r

r = 0.5a

r=0.5\times 3.35\times 10^{-8} cm=1.675\times 10^{-8} cm

1 cm = 10^{10} pm

1.675\times 10^{-8} cm=1.675\times 10^{-8}\times 10^{10}=167.5 pm

Approximate atomic radius for polonium-209 is 167.5 pm.

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