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DiKsa [7]
3 years ago
9

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

ng its initial line of motion with an acceleration of 32.0 m/s2. At this time, its engines fail and the rocket proceeds to move as a projectile. (a) Find the maximum altitude reached by the rocket. 102 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m (b) Find its total time of flight. s (c) Find its horizontal range. 186 Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m
Physics
1 answer:
Serggg [28]3 years ago
5 0

Before the engines fail (0\le t\le3.00\,\rm s), the rocket's horizontal and vertical position in the air are

x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2

y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2

and its velocity vector has components

v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t

v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t

After t=3.00\,\rm s, its position is

x=273\,\rm m

y=362\,\rm m

and the rocket's velocity vector has horizontal and vertical components

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}

After the engine failure (t>3.00\,\rm s), the rocket is in freefall and its position is given by

x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t

y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2

and its velocity vector's components are

v_x=120\,\frac{\rm m}{\rm s}

v_y=159\,\frac{\rm m}{\rm s}-gt

where we take g=9.80\,\frac{\rm m}{\mathrm s^2}.

a. The maximum altitude occurs at the point during which v_y=0:

159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s

At this point, the rocket has an altitude of

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve y=0 for t, then add 3 seconds to this time:

362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s

So the rocket stays in the air for a total of 37.6\,\rm s.

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute x for this time t:

273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m

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3 years ago
Calculate the root mean square velocity of nitrogen molecules at 25 C.
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 root mean square<span>= square root of ( 3RT/M) 

 R = 8.314 J/K/mole 
T = 25 + 273 = 298 K 
M = molecular mas of N2 in kg = 28 X 10^-3 kg 

put values... 

</span><span> root mean square</span> = square root of ( 3 X 8.314 X 298/28 X 10^-3)
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Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the sp
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Answer:

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Explanation:

by conservation of momentum principle we have

m_1v_{i1} + m_2v_{i2} = m_2v_{f2}

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m_2 = \frac{m_1}[\frac{v_f2}{v_{f1}}+1}

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v_{f1} = \sqrt {\frac{(m_1+m_2) v^2_i1}{m_2}

v_{f1} =  v_{i2}\sqrt {\frac{(m_1+m_2)}{m_2}

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substituting this value in above equation to get m2 value

m_2 = \frac{m_1}{\sqrt {\frac{(m_1+m_2)}{m_2}+1}}

solving for m2 we  get

m2 = \frac{m_1}{3}

m_1 = 250 g

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