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.
You can't answer this question because you aren't giving the specific type of seismic waves. There is an s-wave a p-wave and an l-wave. Those are the basic waves. An S-wave cannot travel through a liquid at all. So, obviously it travels slower than any other seismic wave.
<span>It would travel faster because their speed depends on the density and composition of material that they pass through.</span>
It was about 9:30 p.m. sorry if the answer is wrong
Answer:
Work done = = 5 kJ
Explanation:
Given data:
volume of nitrogen
Polytropic exponent n = 1.4
putting all value
polytropic process is given as
work done
= 5 kJ
Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg