Question

Evaluate the power in Watts that a 300-Kg rocket uses to overcome gravity when it travels 5 m at a 5°-angle above the horizontal line in 1.1 s.

Answer 1

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Dear Student,
                     1.s=12×3.2×(32.8)2=1721.344m3. a=v−ut=26.62.47=11.2m/s2s=ut +12at2=18.5×2.47+12×11.2×2.47×2.47=79.78m
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Answer 2

Explanation :

Given that,

Mass of the rocket, m = 300 kg

distance covered, d = 5 m

Time taken, t = 1.1 s

Power is given by rate of doing work. Mathematically, it is written as :

$P=\dfrac{W}{t}$

$P=\dfrac{Fdcos\theta}{t}$

$P=\dfrac{300\times 9.8\times 5\times cos(31)}{1.1}$

$P=\dfrac{300\times 9.8\times 5\times cos(31)}{1.1}$

$P=11454.8\ Watts$

Hence, this is the required solution.