The force of attraction F, between two bodies, varies directly as the product of their masses, \(m_1\) and m\(_2\) and inversely as the square of the distance, d, between them. Given that F = 20N, when m\(_1\) = 25kg, m\(_2\) = 10kg and d = 5m, find:
(I) an expression for F in terms of m\(_1\), m\(_2\) and d;
(ii) the distance, d for F = 30N, m\(_1\) = 7.5kg and m\(_2\) = 4kg
(b) Find the value of x in the diagram
F \(\infty\) \(\frac{m_1m_ 2}{d^2}\)
F = \(\frac{km_1m_2}{d^2}\)
20 = \(\frac{k(25)(10)}{(25)}\)
\(\frac{10k}{10} = \frac{20}{10}\)
k = 2
(i) F = \(\frac{2m_1m_2}{d_2}\)
(ii) Since F = \(\frac{2m_1m_2}{d_2}\)
30 = \(\frac{2(7.5)(4)}{d^2}\)
30 \(\frac{60}{d^2}\)
d\(^2\) = \(\frac{60}{30}\)
d\(^2\) = 2
d = \(\sqrt{2}\)m
(b) Sum of angles = (n - 2) 180
= (5 - 2) 180
= 540\(^o\)
x + x + 20 + x + 60 + x + 40 + x + 80 = 540
5x + 200 = 540
5x = 540 - 200
\(\frac{5x}{5} = \frac{340}{5}\)
x = 68\(^o\)
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}