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3 years ago

a body of mass 5kg falls from height of 10m above the ground what kinetic energy of the body before it strike the ground

1 answer:

4 0

Gravitational Potential Energy (GPE) before fall = Kinetic energy on impact
GPE = mgh
GPE = 5kg x 9.8m/s^2 x 10m
GPE = 490 J
Kinetic Energy = 490 J

(This is assuming that gravitation field strength (g) is 9.8m/s^2, sometimes you may round this to 10m/s^2, therefore making the final result:
Kinetic energy = 500 J)

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$\mathbf{v_a = 4.06 \times 10^7 \ m/s}$

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$\text{The missing part of the question is attached below.} \\ \\ \text{ mass m = 4u and charge q = 2e. Suppose a uranium nucleus with 92 protons decays into }$ $\text{into thorium, with 90 protons, and an alpha particle. The alpha particle is initally at rest at }$ $\text{ the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha}$ $\text{particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest. }$

$\text{So, from the above infromation;}$

$radius (r) = \dfrac{15 \ fm }{2}$

$radius (r) = 7.5 \times 10^{-15} \ m$

$\text{Using the formula for potential energy}$

$U = \dfrac{(9\times 10^9 \ Nm^2/C^2)(2(1.6\times10^{-19} \ C ))(90)(1.6\times 10^{-19} C)}{7.5 \times 10^{-13} \ m}$

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