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.
<h2>The different forces acting on the ball while its in air</h2>
Amy throws a softball through the air. Applied, drag and gravitational forces are acting on the ball while it’s in the air. The softball experiences force as a result of Amy’s throw. As the ball moves, it experiences from the air it passes through.
It also experiences a downward pull because earth has the property to attract everything which is on the earth towards it. The ball is moving in the air but earth applies force on the ball to get back on the ground. Hence, in this way, gravitational force applies.
There is also a drag force which results due to friction that is present in the air. It resist to move ball in the air and there will also be applied force which is given by a person who throws by applying force.
Answer:
128.21 m
Explanation:
The following data were obtained from the question:
Initial temperature (θ₁) = 4 °C
Final temperature (θ₂) = 43 °C
Change in length (ΔL) = 8.5 cm
Coefficient of linear expansion (α) = 17×10¯⁶ K¯¹)
Original length (L₁) =.?
The original length can be obtained as follow:
α = ΔL / L₁(θ₂ – θ₁)
17×10¯⁶ = 8.5 / L₁(43 – 4)
17×10¯⁶ = 8.5 / L₁(39)
17×10¯⁶ = 8.5 / 39L₁
Cross multiply
17×10¯⁶ × 39L₁ = 8.5
6.63×10¯⁴ L₁ = 8.5
Divide both side by 6.63×10¯⁴
L₁ = 8.5 / 6.63×10¯⁴
L₁ = 12820.51 cm
Finally, we shall convert 12820.51 cm to metre (m). This can be obtained as follow:
100 cm = 1 m
Therefore,
12820.51 cm = 12820.51 cm × 1 m / 100 cm
12820.51 cm = 128.21 m
Thus, the original length of the wire is 128.21 m
