Answer:
-54 m/s²
Explanation:
Acceleration is defined as the change in velocity of a body with respect to time. Mathematically,
Acceleration A = change in velocity/time
A = dv/dt
Given Vx = at − bt³
The time at which the particle reaches its maximum displacement is at when vx = 0.
0 = at-bt³
t(a-bt²) = 0
a-bt² = 0
a = bt²
t² = a/b ... (1)
A = dvx/dt = a - 3bt²(by differentiating)
Acceleration = a - 3bt²... (2)
Substituting t² = a/b into equation 2 will give;
Acceleration = a - 3b(a/b)
Acceleration = a-3a
Acceleration = -2a
Substituting the value of a = 27m/s into the resulting equation of acceleration gives;
Acceleration = -2(27)
Acceleration = -54m/s²
Therefore at maximum displacement in the positive x direction, the acceleration of the particle will be -54m/s²
Answer:
U can walk to the bathroom
Explanation:
Answer: 
Explanation:
Given
The dimension of the plate is 
The gap between the plate is 
Voltage applied 
The capacitance of the capacitor is

Charge acquired by the capacitor

Complete Question:
Two 3.0µC charges lie on the x-axis, one at the origin and the other at 2.0m. A third point is located at 6.0m. What is the potential at this third point relative to infinity? (The value of k is 9.0*10^9 N.m^2/C^2)
Answer:
The potential due to these charges is 11250 V
Explanation:
Potential V is given as;

where;
K is coulomb's constant = 9x10⁹ N.m²/C²
r is the distance of the charge
q is the magnitude of the charge
The first charge located at the origin, is 6.0 m from the third charge; the potential at this point is:

The second charge located at 2.0 m, is 4.0 m from the third charge; the potential at this point is:

Total potential due to this charges = 4500 V + 6750 V = 11250 V