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: beta particles
Explanation:
Given mass = 14.0 g
Molar mass = 137 g/mol
According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number of particles.
1 mole of cesium contains atoms =
0.102 moles of cesium contains atoms =
The relation of atoms with time for radioactivbe decay is:
Where =atoms left undecayed
= initial atoms
t = time taken for decay = 3 minutes
= half life = 30.0 years = minutes
The fraction that decays :
Amount of particles that decay is =
Thus beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.
Answer:
g = 1.64m/s²
Explanation:
1.5m in 0.078s
V = 15 / 0.078
= 19.23m/s
Tension = mg
μ = 3.10 × 10⁻⁴
T = V²μ
mg = V²μ
g = V²μ / m
g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)
g = 1.64m/s²
Answer:
The answer is
<h2>270 m</h2>
Explanation:
To find the distance when given the velocity and time we use the formula
<h3>distance = velocity × time</h3>
From the question
velocity of the ball = 18 m/s
time = 15 s
So the distance is
distance = 18 × 15
We have the final answer as
<h3>270 m</h3>
Hope this helps you