A person in kayak stars padding and it accelerates from 0 to

A person in kayak stars padding, and it accelerates from 0 to 0.44 m/s in a distance of 0.68 m. If the combine magnitude of the net force on the kayak ? A rocket of mass 5.5times 105 kg is in fight its thrust is directed of an angle of 55.0^8 above the horizontal and a magnitude of 6.7 times 106 N. Find the magnitude of the rocket acceleration. On earth two parts as a space weight 15000 N and 4600N. These parts are separated by a center to center distance of 9.0m and may be treated as uniform spherical objects. Find the magnitude of the gravitational force that each part exerts on the other out in space, far from any other object. 7.5times10^-1N 7.6 times 10^5N 5.7times 10^-4N 5.6times 10^6N

Solution

2) V^2-U^2= 2as from euqtion of motion.

so v=.44 m/s . u=0 s= .68m

so a=.44*.44/(2*.68) m/s^2

F=ma = 91*.44*.44/(2*.68) N = 12.95 N

3)

8.70 x 10^6 N is acting 55.0° above the horizontal .
The vertical component upwards = 8.70 x 10^6 x sin 55.0* = 7126622 N
Wt. vertically downwards = 9.81 x 5.5 x 10^5 = 5401000 N

Net force acting upwards = 7126622- 5401000 = 1725622 N

Horizontal component of 8.70 x 10^6 N is acting 55.0° above the horizontal .
= 8.70 x 10^6 x cos 55.0* = 4990114 N

Resultant force = ( 1725622^2 + 4990114^2) ^0.5 = 5280057 N
F / m = acceleration = 5280057 / 5.5 x 10^5 = 9.6 m / s^2

3) F=Gm1m2/r^2 = 6.674 *10^-11*15000*4600/(9*9*9.8*9.8) =5.7 *10^-5 N