Answer:
F = 58.1N
θ = 10.4° South of East.
Explanation:
Let the situation be as represented in the diagram attached. To determine the net force acting on the bear, we need to resolve the forces into component forms.
Taking west-east as the +x axis and south-north as +y axis.
F1x = 0N (force is directed due north and so has not x component)
F2x = – F2cos30° = –30cos30° = –26N
F3x = – 25N (force is directed due west and so has only x component)
F4x = –F4sin10° = -35sin10° = –6.08N
Fx = F1x + F2x + F3x + F4x =0 – 26 –25 – 6.08 = –57.08N
F1y = 20N (force is directed due north and so has not x component)
F2y = F2sin30° = 30sin30° = 15N
F3y = 0N (force is directed due west and so has only x component)
Fy = –F4cos10° = -35cos10° = –34.5N
Fy = F1y + F2y + F3y + F4y = 20 + 15 + 0 –34.5 = –10.5N
F = √(Fx² + Fy²)
F = √((-57.1² + (-10.5)²) = 58.1N
θ = tan-¹(Fy/Fx) = tan-¹(-10.5/-57.1) = 10.4° South of East.
F = 58.1N
θ = 10.4° South of East.
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
Probably because of the drag coefficient and the density of the liquid.
