(a) 3675 N
Assuming that the acceleration of the rocket is in the horizontal direction, we can use Newton's second law to solve this part:
where
is the horizontal component of the force
m is the mass of the passenger
is the horizontal component of the acceleration
Here we have
m = 75.0 kg
Substituting,
(b) 3748 N, 11.3 degrees above horizontal
In this part, we also have to take into account the forces acting along the vertical direction. In fact, the seat exerts a reaction force (R) which is equal in magnitude and opposite in direction to the weight of the passenger:
where we used
as acceleration of gravity.
So, this is the vertical component of the force exerted by the seat on the passenger:
and therefore the magnitude of the net force is
And the direction is given by
Answer:
A conduction is the answer
Explanation:
Answer: Velocity terminal = 0.093m/s
Explanation:
1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)
= (0.0604/2 - 0.06/2)m
= 2×10^-4
Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L
= (π×0.06×0.4)m²
= 0.075m²
Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.
Shearing stress = u×V.terminal/h = 0.86×V/0.0002
= 4300Vterminal
Therefore, Fw = shearing stress × A
30N = 4300Vterminal × 0.075
V. terminal = 30/4300 m.s
V. terminal = 0.093m/s
