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borishaifa [10]
3 years ago
6

How do astronomers use rocks from the moon to estimate the age of the solar system?

Physics
1 answer:
Firlakuza [10]3 years ago
6 0

Answer:

Might be better to use geophysicists to date the moon rock via   Radioactive Age-Dating  Carbon 12

Explanation:

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<span>Kirchhoff's laws apply to AC circuits to either two cases: instantaneousneous values of currents and voltages or to complex values of currents and voltages. However, this never applies to: rms values of currents and voltages. Kirchoff's law relates to the current, voltage and resistance to multiple nodes</span>
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Bill and Amy want to ride their bikes from their neighborhood to school which is 14.4 kilometers
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First we should convert 14.4 km into meters using the conversion factor 1km = 1000m; thus, 14.4 km = 14,400 m. Next, we should convert all minutes into seconds <span>using the conversion factor 1 min = 60 seconds; thus, 40 mins = 2400 seconds while 20 minutes = 1200 seconds. 
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4 0
3 years ago
A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con
Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5  x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \  \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm

The magnitude of the electric field is now given as;

E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

5 0
3 years ago
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