Answer:
d = 2021.6 km
Explanation:
We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them
Airplane 1
Height y₁ = 800m
Angle θ = 25°
cos 25 = x / r
sin 25 = z / r
x₁ = r cos 20
z₁ = r sin 25
x₁ = 18 103 cos 25 = 16,314 103 m
= 16314 m
z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m
2 plane
Height y₂ = 1100 m
Angle θ = 20°
x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m
z₂ = 20 103 without 25 = 8.452 103 m = 8452 m
The distance between the planes using the Pythagorean Theorem is
d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2
Let's calculate
d² = (18126-16314)² + (1100-800)² + (8452-7607)²
d² = 3,283 106 +9 104 + 7,140 105
d² = (328.3 + 9 + 71.40) 10⁴
d = √(408.7 10⁴)
d = 20,216 10² m
d = 2021.6 km
The correct answer to the question is that the lost mass has been converted into energy.
EXPLANATION:
From Einstein's theory, we know that energy and mass are inter convertible .
When some amount of mass is lost, same amount of energy equivalent to mass is produced.
Let us consider m is the mass lost during any reaction. Hence, the amount of energy produced will be-
Energy E =
Here, c is the velocity of light i.e c = 
As per the question, uranium-235 undergoes fission. The amount of mass defect is 0.1 %.
The mass defect is defined as the difference between mass of reactants and products. During the fission, energy is produced.
The energy produced in this reaction is nothing else than the energy equivalent to mass defect. Approximately 199.5 Mev of energy equivalent to this mass defect is produced in this reaction.
Change in velocity = d(v)
d(v) = v2 - v1 where v1 = initial speed, v2 = final speed
v1 = 28.0 m/s to the right
v2 = 0.00 m/s
d(v) = (0 - 28)m/s = -28 m/s to the right
Change in time = d(t)
d(t) = t2 - t1 where t1 = initial elapsed time, t2 = final elapsed time
t1 = 0.00 s
t2 = 5.00 s
d(t) = (5.00 - 0.00)s = 5.00s
Average acceleration = d(v) / d(t)
(-28.0 m/s) / (5.00 s)
(-28.0 m)/s * 1 / (5.00 s) = -5.60 m/s² to the right
At r = 2R> R The expression for the electric field will be given by: (2R)^2*E=kQ. Where, k=(9*10^9)N.m/C^2, Q=(8*10^-10)C and R=0.025m. So substituting and clearing, we have that the magnitude of the electric field will be: E=(9*10^9)*(8*10^-10)/((2*0.025)^2)=2880 N / C.