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4vir4ik [10]
3 years ago
13

a model rocket is launched at an angle of 55 to the horizontal with a momentum of 20 kg-m/. If the vertical component of its vel

ocity is 40 m/s what is the mass of the rocket?
Physics
1 answer:
den301095 [7]3 years ago
4 0
Given:\\ \alpha =55^\circ\\p=20kg\cdot  \frac{m}{s} \\v_v=40 \frac{m}{s} \\\\Find:\\m=?\\\\Solution:\\\\p_v=mv_v\Rightarrow m= \frac{p_v}{v_v} \\\\ \frac{p_v}{p} =\cos  \alpha \Rightarrow p_v=p\cos  \alpha \\\\m= \frac{p\cos  \alpha }{v_v} \\\\m= \frac{20kg\cdot  \frac{m}{s} \cdot \cos 55^\circ }{40 \frac{m}{s} } \approx 0.287kg=287g
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Two boxes are connected to each other by a string as shown in the figure. The 10-n box slides without friction on the horizontal
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Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface
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Explanation:

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                   R = 2r

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                      = 0.2 m

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As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, \frac{U}{2}.

Hence,   K.E = \frac{U}{2}

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Now, we will calculate the speed of balls as follows.

                 V = \sqrt{\sqrt{\frac{2 \times K.E}{m}}

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Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

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