In Newton's cannonball experiment, if the velocity is equal the orbital velocity the Cannon ball
1 answer:
You might be interested in
Answer:
minimum length of runway is needed for take off 243.16 m
Explanation:
Given the data in the question;
mass of glider = 700 kg
Resisting force = 3700 N one one glider
Total resisting force on both glider = 2 × 3700 N = 7400 N
maximum allowed tension = 12000 N
from the image below, as we consider both gliders as a system
Equation force in x-direction
2ma = T -f
a = T-f / 2m
we substitute
a = (12000 - 7400 ) / (2 × 700 )
a = 4600/1400
a = 3.29 m/s²
Now, let Vf be the final speed and Ui = 0 ( as starts from rest )
Vf² = Ui² + 2as
solve for s
Vf² = 0 + 2as
2as = Vf²
s = Vf² / 2a
given that take of speed for the gliders and the plane is 40 m/s
we substitute
s = (40)² / 2×3.29
s = 1600 / 6.58
s = 243.16 m
Therefore, minimum length of runway is needed for take off 243.16 m
Fission reaction is given
¹₀n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3(¹₀n)
The atomic masses of the fission products are 97.912735 u for ⁹⁸₄₀Zr and 134.916450 u for ¹³⁵₅₂Te. Hence, the energy released in fission reaction is 191.715 MeV.
How to find the energy released in the given fission reaction?
We know, that the atomic mass of the elements are as follows:
- ⁹⁸₄₀Zr - 97.9120u
- ¹³⁵₅₂Te - 134.9087u
- ²³⁵₉₂U - 235.0483923u
- n = 1.008665u
In order to find the mass difference, we will calculate the initial mass and the mass of products.
The equation for the reaction:
n + ²³⁵₉₂U → ⁹⁸₄₀Zr + ¹³⁵₅₂Te +3n
Mass of initial reagents is 1.008665u + 235.0483923u = 233.052588u
Product of the reagents is 97.9120u + 134.9087u + 3(1.008665u) = 235.846773u
Now, using the formula of
Hence, the energy released in fission reaction is 191.715 MeV.
To learn more about fission reactor, refer to:
#SPJ4