We Know, F = m.a
Here, m = 2 Kg
& a = 2 m/s²
Substitute for it,
F = 2*2 Kgm/s²
F = 4 Kgm/s²
Final velocity = 0m/s
Initial velocity = 38m/s
Acceleration = 0-38/6 = -38/6 = -6.33m/s2
Input current passing easy to can you make work output of a machine greater than the work
Answer:
The distance is shortenend by factor .1715
Explanation:
5 n = 1/r^2
sqrt (1/5) = r
170 n = 1 / ( x sqrt(1/5))^2
(xsqrt 1/5)^2 = 1/170
x sqrt 1/5 = .076696
x = .1715
