Given v_in = 20 m/s and a = 3 m/s2, assuming that the body
moves at constant acceleration, the motion is modeled by the equation:
s(t) = (v_in)t + (1/2)a(t^2)
where s(t) is the distance traveled
substituting the given,
s(t) = 20t + (3/2)(t^2)
at t = 3
s(t) = 20(3) + (3/2)(3)^2
= 73.5 m
Answer:
If the height of the ramp is changed at a different height then the speed of the car will change.
Answer: T = 93 N
Explanation:
Assuming the pulley is ideal meaning frictionless as mentioned and also negligible mass.
ΣF = Σma
Mg - mg = Ma + ma
a = g(M - m) / (M + m)
Now looking only at the larger mass as it falls
Mg - T = Ma
T = Mg - Ma
T = Mg - Mg(M - m) / (M + m)
T = Mg(1 -(M - m) / (M + m))
T = 16(9.8)(1 - (16 - 6.7) / (16 + 6.7))
T = 93 N
or looking only at the smaller mass
T - mg = ma
T = m(g + a)
T = m(g + g(M - m) / (M + m))
T = mg(1 + (M - m) / (M + m))
T = 6.7(9.8)(1 + (16 - 6.7) / (16 + 6.7))
T = 93 N
Answer:
An object at rest is described by Newton's First Law of Motion. An object in static equilibrium has zero net force acting upon it. The First Condition of Equilibrium is that the vector sum of all the forces acting on a body vanishes.
Explanation: