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Answer:
The acceleration required by the rocket in order to have a zero speed on touchdown is 19.96m/s²
The rocket's motion for analysis sake is divided into two phases.
Phase 1: the free fall motion of the rocket from the height 2.59*102m to a height 86.9m
Phase 2: the motion of the rocket due to the acceleration of the rocket also from the height 86.9m to the point of touchdown y = 0m.
Explanation:
The initial velocity of the rocket is 0m/s when it started falling from rest under free fall. g = 9.8m/s² t1 is the time taken for phase 1 and t2 is the time taken for phase2.
The final velocity under free fall becomes the initial velocity for the accelerated motion of the rocket in phase 2 and the final velocity or speed in phase 2 is equal to zero.
The detailed step by step solution to the problems can be found in the attachment below.
Thank you and I hope this solution is helpful to you. Good luck.
Answer:
6.5 x 10^32 eV
Explanation:
mass of particle, mo = 1 g = 0.001 kg
velocity of particle, v = half of velocity of light = c / 2
c = 3 x 10^8 m/s
Energy associated to the particle
E = γ mo c^2
Convert Joule into eV
1 eV = 1.6 x 10^-19 J
So,
Answer:
The actual angle is 30°
Explanation:
<h2>Equation of projectile:</h2><h2>y axis:</h2>the velocity is Zero when the projectile reach in the maximum altitude:
When the time is vo/g the projectile are in the middle of the range.
<h2>x axis:</h2>R=Range
**sin(2A)=2sin(A)cos(A)
<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>Let B the actual angle of projectile
2B=60°
B=30°