Answer:
Maximum height reached by the rocket is

total time of the motion of rocket is given as

Explanation:
Initial speed of the rocket is given as

acceleration of the rocket is given as

engine stops at height h = 150 m
so the final speed of the rocket at this height is given as



so maximum height reached by the rocket is given as the height where its final speed becomes zero
so we will have



Now the total time of the motion of rocket is given as
1) time to reach the height of 150 m



2) time to reach ground from this height



so total time of the motion of rocket is given as

Bergeron–Findeisen Process.
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
Answer:
30 T
Explanation:
The magnetic field due to first wire at point A is 
Magnetic field due to second wire at point A is 
As the current flows in opposite direction in both the wire so net magnetic magnetic field at point A is 
So net magnetic field
as
is given as 10 T in question