We are given with
l = 8.00 m
θ = 37.0°
m = 470 kg
r = 0.300 m
μ = 0.120
To solve the magnitude of the force required to roll the cylinders up the ramp, we do a force balance
F = Ff
F = μFn
F = mgμ sin θ
substituting the given values
F = 470 (9.81) (0.120) sin 37.0°
F = 333.97 N<span />
Distance=2m
because 200cm = 2m
so 4m-2m=2m
Answer:
v = 30.39 m/s
Explanation:
given,
mass of glider,M = 680 Kg
mass of the skydiver, m = 68 Kg
horizontal velocity,V = 30 m/s
when skydiver releases, velocity,v = 30 m/s
velocity if the glider,v' = ?
use the conservation of momentum
M V = m' v' + m v
m' = 680-68 = 612 Kg
680 x 30 = 612 x v + 60 x 30
612 v = 18600
v = 30.39 m/s
since the skydiver's speed will be the same as before release
The glider should continue to travel at 30.39 m/s since there are no external forces acting on it
Answer:
e^-4t e^-5t
Explanation:
solving s²+9s+20 quadratically we have (s+4)(s+5)
x(s) can be written as x(s) =(1/s+4)(1/s+5)
if we take the laplace inversve
L-¹ (s)=L-1(1/s+4) L-1(1/s+5)
we have e^-4t e^-5t
