At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.
Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.
If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
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92 percent because you multiply it by 2
Answer:
a) technician A
Explanation: A parking brake is a secondary and support brake system which is usually applied when the vehicle is not in motion,this is done to prevent any form of motion and the possibility of any form of accident.
A secondary circuit is a circuit or coil in which current is generated through induction from another circuit or coil called the primary circuit,it has not been discovered that a low pressure from that circuit has led to poor trailer parking brake performance.
Answer:
The correct option is A = 1960 N/m²
Explanation:
Given that,
Mass m= 20,000kg
Area A = 100m²
Pressure different between top and bottom
Assume the plane has reached a cruising altitude and is not changing elevation. Then sum the forces in the vertical direction is given as
∑Fy = Wp + FL = 0
where
Wp = is the weight of the plane, and
FL is the lift pushing up on the plane.
Let solve for FL since the mass of the plane is given:
Wp + FL = 0
FL = -Wp
FL = -mg
FL = -20,000× -9.81
FL = 196,200N
FL should be positive since it is opposing the weight of the plane.
Let Equate FL to the pressure differential multiplied by the area of the wings:
FL = (Pb −Pt)⋅A
where Pb and Pt are the static pressures on bottom and top of the wings, respectively
FL = ∆P • A
∆P = FL/A
∆P = 196,200 / 100
∆P = 1962 N/m²
∆P ≈ 1960 N/m²
The pressure difference between the top and bottom surface of each wing when the airplane is in flight at a constant altitude is approximately 1960 N/m². Option A is correct
