I think in parallel circuits.
Answer: A is Compression and B is Rarefaction.
Explanation:
i think it's right. hope it helps.
1) 14g
2) 24 fl oz
3)10 cups
4)12 and 1/2 cups
Answer:
k = 1.30
Explanation:
For this exercise let's write the capacitance in air and with dielectric
air C₀ = Q / DV
dielectric C = k Q / DV
They tell us that the capacitor is charged and then the battery is disconnected, therefore the charge stored on the plate remains constant.
therefore the capacitance a changes to the value
C = k C₀
The voltage in the presence of dielectric must meet the relationship
ΔV = ΔV₀ / k
k = ΔV₀ /ΔV
let's calculate
k = 60/46
k = 1.30
Answer:
Explanation:
Electric field at the surface of the the lead 208 = KQ/ R²
where K = 8.99 × 10⁹ Nm² /C²
Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C
V of the lead = 208 v of a proton assuming they both are sphere
4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton
R³ = 208 r³
R = ∛( 208 r³) = 5.92r
replace r with 1.20 x 10-15 m
R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m
E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² = 0.233 × 10²² N/C = 2.33 × 10²¹ N/C
