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lorasvet [3.4K]
3 years ago
14

A basketball weighting 0.5kg is flying through the air at 10m/s. what is the kinetic energy

Physics
1 answer:
jolli1 [7]3 years ago
4 0
Kinetic energy = (mv^2)/2. So (.5)(10^2)/2 = 25 J
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Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

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According to the lecture, all of the following are defense mechanisms except __________.
Liula [17]
<span>According to the lecture, all of the following are defense mechanisms except: Moderate Exercise

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A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises b
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The work done by gas is 0.753 J and change in internal energy is 4.247J

So we are given that mass is 2kg , radius 1 cm and the amount of heat is 5 cm

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Here pressure is 10^5 pa

So W = 10^5\times7.53\times10^-(6)

Therefore W = 0.753 J

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Learn more about Work done here

brainly.com/question/16951089

#SPJ1  

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