The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.
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
The answer might be C ? hope it's right
Explanation:
Work cannot be increased by using a machine of some kind.
Easy !
Take any musical instrument with strings ... a violin, a guitar, etc.
The length of the vibrating part of the strings doesn't change ...
it's the distance from the 'bridge' to the 'nut'.
Pluck any string. Then, slightly twist the tuning peg for that string,
and pluck the string again.
Twisting the peg only changed the string's tension; the length
couldn't change.
-- If you twisted the peg in the direction that made the string slightly
tighter, then your second pluck had a higher pitch than your first one.
-- If you twisted the peg in the direction that made the string slightly
looser, then your second pluck had a lower pitch than the first one.
