Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
From the conservation of linear momentum of closed system,
Initial momentum = final momentum
Mass of the student, M = 59 kg
Mass of the laser boat, m = 42 kg
Initial speed of student + laser boat, u =0
Final speed of laser boat, v = 1.5 m/s
Final speed of the student = V
(M+m) u =M V +m v
0 = (59 kg) V + (42 kg) (1.5m/s)
V = - 1.06 m/s
Thus, the speed of the student is 1.06 m/s in the opposite direction of the motion of boat.
Answer:
The velocity of the man is 0.144 m/s
Explanation:
This is a case of conservation of momentum.
The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.
Mass of ball = 0.65 kg
Mass of the man = 54 kg
Velocity of the ball = 12.1 m/s
Before collision, momentum of the ball = mass x velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After collision the momentum of the man and ball system is
(0.65 + 54)Vf = 54.65Vf
Where Vf is their final common velocity.
Equating the initial and final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
a. FTh = 30 N
b. Fw = 30 N
c. a = 200 m/s2
Explanation:
See full explanation in the picture. Please rate as brainliest
