Answer:
vB = 15.4 m/s
Explanation:
Principle of conservation of energy:
Because there is no friction the mechanical energy is conserve
ΔE = 0
ΔE : mechanical energy change (J)
K : Kinetic energy (J)
U: Potential energy (J)
K = (1/2)mv²
U = m*g*h
Where :
m: mass (kg)
v : speed (m/s)
h : hight (m)
Ef - Ei = 0
(K+U)final - (K+U)initial =0
(K+U)final = (K+U)initial
((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:
((1/2)vB² + g*hB = (1/2 )vA²+ g*hA
(1/2) (vB)² + (9.8)*(14.7) = 0 + (9.8)(26.8 )
(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)
(vB)² = (2)(9.8)(26.8 - 14.7)
(vB)² = 237.16

vB = 15.4 m/s : speed of the cart at B
u= 215 km/hr = 215 * 1000/ 3600 = aprx 60m/s
v=0
t=2.7sec
v= u - at
u= at
60/2.7 = 22.23 m/s^2
Hope it helps
Trick question? In order to have kinetic energy, an object must be moving. Therefore, in this case, kinetic energy would be 0. If it were asking about potential energy, it would be a different story.
Answer:
hello your question lacks some data and required diagram
G = 77 GPa, т all = 80 MPa
answer : required diameter = 252.65 * 10-^3 m
Explanation:
Given data :
force ( P ) = 660 -N force
displacement = 15 mm
G = 77 GPa
т all = 80 MPa
i) Determine the required diameter of shaft BC
considering the vertical displacement ( looking at handle DC from free body diagram )
D' = 0.3 sin∅ , where D = 0.015
hence ∅ = 2.8659°
calculate the torque acting at angle ∅ of CD on the shaft BC
Torque = 660 * 0.3 cos∅
= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N
hello attached is the remaining part of the solution
Newton's Second law of motion:
Force = (mass) x (acceleration)
Force = (15kg) x (8m/s²) = 120 kg-m/s² = 120 newtons