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Nata [24]
3 years ago
13

Please help due is 1 hour please its very urgent :(((

Physics
1 answer:
Tasya [4]3 years ago
8 0

Answer:

fnet = ma

fnet = (person + sled)*a

a = fapp / person + sled

= 300 N / 70kg

= 4.3 m/s^2

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Distance of closest approach, r=1.91\times 10^{-14}\ m

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The distance of closest approach for two charged particle is given by :

r=\dfrac{k2e^2(m_p+m_a)}{2m_am_pv_p^2}

r=\dfrac{9\times 10^9\times 2(1.6\times 10^{-19})^2(1.67\times 10^{-27}+6.68\times 10^{-27})}{2\times 6.68\times 10^{-27}\times 1.67\times 10^{-27}(0.01\times 3\times 10^8)^2}

r=1.91\times 10^{-14}\ m

So, their distance of closest approach, as measured between their centers 1.91\times 10^{-14}\ m. Hence, this is the required solution.

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