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Alexandra [31]
3 years ago
5

An experimental rocket designed to land upright falls freely from a height of 2.59 102 m, starting at rest. At a height of 86.9

m, the rocket's engines start and provide constant upward acceleration until the rocket lands. What acceleration is required if the speed on touchdown is to be zero?'

Physics
1 answer:
aleksandr82 [10.1K]3 years ago
5 0

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.

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3 years ago
Sam is pulling a box up to the second story of his apartment via a string. The box weighs 92.6 kg and starts from rest on the gr
grin007 [14]

Answer: 2.7 seconds

Explanation:

We only want to answer: How long until the box reaches a height of 13.1 m?

Then we only must integrate the movement equations.

We know that the velocity of the box increases by 3.6 m/s every second, then the acceleration is constant, and can be written as:

a(t) = 3.6m/s^2

Now, for the velocity, we should integrate over time, and because we know that the box starts from rest, the initial velocity (the constant of integration) will be zero.

v(t) = (3.6m/s^2)*t

For the position equation we should integrate again over time, and if we define the position 0 as the ground, we know that the box starts at the ground, then the initial position (the constant of integration) will be zero.

p(t) = (1/2)*(3.6m/s^2)*t^2.

Now we want to find how long will take until the height of the box is equal to 13.1m

Then we must solve:

p(t) = 13.1m =  (1/2)*(3.6m/s^2)*t^2

Let's solve this for t.

13.1m =  (1/2)*(3.6m/s^2)*t^2

13.1m*2 = (3.6m/s^2)*t^2

26.2m/(3.6m/s^2) = t^2

7.27<u>7</u>.... s^2 = t^2

√(7.27<u>7</u>.... s^2) = t = 2.7 seconds.

So it will take 2.7 seconds for the box to reach the height of 13.1m

4 0
3 years ago
The rotational inertia of a collapsing spinning star changes to 1/6 its initial value. What is the ratio of the new rotational k
storchak [24]

Answer:

Explanation:

Let the initial rotational inertia be I and final rotational inertia be I / 6 .

Let the initial angular velocity be ω₁ and final angular velocity be ω₂.

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= ( 1 / 6 ) x ω₂ ² / ω₁ ²

=  ω₂ ² / 6ω₁ ²

= 36 ω₁ ² / 6ω₁ ²

= 6 .

6 0
3 years ago
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