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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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When a material absorbs energy from light the energy causes the material to do what?
boyakko [2]

Answer:

To change.

Explanation:

The Law of Conservation of energy states that energy cannot be destroyed but can only be transformed or converted from one form to another. For example, light energy to heat energy.

When a material absorbs energy from light the energy causes the material to change.

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3 years ago
The venn diagram shown below compares the nuclear reactions in the sun and nuclear power plants.
BARSIC [14]

Answer:

Formation of new elements

Explanation:

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3 years ago
A pool ball of 0.17 kg moves at 10 m/s and strikes a stationary ball (at rest) also at 0.17 kg.
Elenna [48]

The final velocity of the second ball is +10 m/s

Explanation:

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system must be conserved before and after the collision, so we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

where:

m_1 = 0.17 kg is the mass of the first ball

u_1 = 10 m/s is the initial velocity of the first ball (we take its direction as positive direction)

v_1 = 0 is the final velocity of the first ball

m_2 = 0.17 kg is the mass of the second ball

u_2 = 0 is the initial velocity of the second ball

v_2 is the final velocity of the second ball

Re-arranging the equation and substituting the values, we find:

v_2 = \frac{m_1 u_1}{m_2}=\frac{(0.17)(10)}{0.17}=10 m/s

And since the sign is positive, the direction is the same as the initial direction of the first ball.

Learn more about momentum here:

brainly.com/question/7973509

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5 0
4 years ago
A light ball is dropped straight down off a ledge and hits the ground. A heavy ball is thrown straight down from the ledge, rele
Anettt [7]

Answer:

(d) I and III

Explanation:

Motion of the 1st  light ball

s = gt_1^2/2

s = v_0t_2 + gt_2^2/2

I. Since the heavier starts with an initial velocity v0, and both balls end up traveling the same distance s. From the 2 motion equation above we can conclude that their time must be different

II. Both balls are affected by the same gravitational acceleration g

III. Velocity before the impact of the light ball

v_1^2 = 2gs

And the heavy ball

v_2^2 - v_0^2 = 2gs

Since they have the same acceleration g and distance s. But the heavy ball has an initial push. It would end up with a larger speed.

7 0
3 years ago
One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at
Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature(T_i)=300 K

P_i=10 atm

P_f=2 atm

Work done in iso-thermal process=P_iV_iln\frac{P_i}{P_f}

P_i=initial pressure

P_f=Final Pressure

W=10\times 2.463\times ln\frac{10}{2}=39.64 J

Since it is a iso-thermal process therefore q=w

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(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is=P\Delta V

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V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L

\Delta W=1\times (12.315-2.463)=9.852 J

\Delta q=\Delta W=9.852 J

\Delta U=0

8 0
4 years ago
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