Answer: High pass filters, and in particular LC high pass filters are used in many RF applications where they block the lower frequencies and allow through higher frequency signals. Typically LC filters are used for the higher radio frequencies where active filters are not so manageable, and inductors more appropriate.
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Answer:
Explanation:
When an object is heated then it becomes brighter and bluish compared to the initial condition. This happens because when an object is given heat then the electron in the ground state gets excited and reaches some higher state. After reaching a higher state electron make the transition to lower state and simultaneously exhibit the color which is visible with naked eyes.
Answer:
Explanation:
If the passage of the waves is one crest every 2.5 seconds, then that is the frequency of the wave, f.
The distance between the 2 crests (or troughs) is the wavelength, λ.
We want the velocity of this wave. The equation that relates these 3 things is
and filling in:
so
v = 2.5(2.0) and
v = 5.0 m/s
Answer:
3.135 kN/C
Explanation:
The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}
Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
So, E = qz/{4πε₀[√(z² + R²)]³}
E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}
E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}
E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}
E = 6.839 × 10³ Cm²/[1.297 m]³F
E = 6.839 × 10³ Cm²/2.182 m³F
E = 3.135 × 10³ V/m
E = 3.135 × 10³ N/C
E = 3.135 kN/C
