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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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Answer with Explanation:

We are given that

Mass of squirrel,m=530 g=\frac{530}{1000}=0.530 kg

1kg=1000 g

Area=A=935 cm^2=935\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

Height,h=4.4 m

C=1

\rho=1.21 kg/m^3

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Area=l\times b=0.116\times 0.232=0.0269 m^2

Terminal velocity,v_t=\sqrt{\frac{2mg}{\rho CA}}

Where g=9.8 m/s^2

Using the formula

v_t=\sqrt{\frac{2\times 0.530\times 9.8}{1.21\times 1\times 0.0269}}

v_t=17.86 m/s

The velocity of person,v=\sqrt{2gh}

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Use the motion map to answer the question.
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Answer:

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Natasha_Volkova [10]

Answer:

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Explanation:

The potential energy is given by:

U=Q*V

where:

Q is the charge

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Potential Difference=V=\frac{kq}{r}

So,

Potential\ Energy=\frac{Qkq}{r} \\Q=q\\Potential\ Energy=\frac{kq^2}{r}

Where:

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For 3 Electrons Potential Energy or work Done is:

Potential\ Energy=3*\frac{kq^2}{r}

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Answer:

Answer is

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B. Yes

C. No

Refer below.

Explanation:

Refer to the picture for brief explanation.

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3 years ago
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