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polet [3.4K]
2 years ago
10

A 3kg object has an initial velocity (6i - 2j) m/s (a ) what is its kinetic energy at this time? (b) Find total work done on the

object as its velocity changes to (8i+4j) m/s​?
Physics
1 answer:
guapka [62]2 years ago
3 0

Answer:

K.E =  \frac{1}{2} m {v}^{2}  \\  {v}^{2}_i  =  {v}^{2} _x + {v}^{2} _y \\  =  {(6)}^{2}  +  {( - 2)}^{2}  = 36 + 4 = 40m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (40) = 60J \\  \\ {v}^{2}_f  =  {v}^{2} _x + {v}^{2} _y \\  =  {(8)}^{2}  +  {(4)}^{2}  = 80m. {s}^{ - 1} \\ K.E_i =  \frac{1}{2} (3) (80) = 120J \\ W_{net}=K.E_f-K.E_i \\  = 120J - 60J \\  = 60J

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Answer:

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Explanation:

Parameters given:

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A) The distance traveled by the planet is half of the circumference of the orbit (which is circular).

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B) Work done is given as:

W = F * D

W = 652 * 10²² * 1.88 * 10¹¹

W = 1.22 * 10³⁴ J

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K. E. = 0.5 * 7 * 10^25 * (2.36 * 10^4)²

K. E. = 1.95 * 10³⁴ J

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Wavelength is the distance between 2 adjacent points in a wave

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