Answer:
(A) Q = 321.1C (B) I = 42.8A
Explanation:
(a)Given I = 55A−(0.65A/s2)t²
I = dQ/dt
dQ = I×dt
To get an expression for Q we integrate with respect to t.
So Q = ∫I×dt =∫[55−(0.65)t²]dt
Q = [55t – 0.65/3×t³]
Q between t=0 and t= 7.5s
Q = [55×(7.5 – 0) – 0.65/3(7.5³– 0³)]
Q = 321.1C
(b) For a constant current I in the same time interval
I = Q/t = 321.1/7.5 = 42.8A.
What is he minumum coating of thickness needed to ensure that lifght of waveelntght 5660 mbnd si
I,think potential energy is mgh so 65*100*9,81
The AREA of the shaded region is the moving object's displacement.
Answer:
The velocity of the motorboat after 6s is 24 m/s.
Explanation:
Given;
acceleration of the motorboat, a = 4.0 m/s²
initial velocity of the motorboat, u = 0
time of motion of the motorboat = 6s
Apply the following kinematic equation to determine the velocity of the motorboat after 6 ;
v = u + at
v = 0 + (4 x 6)
v = 24 m/s
Therefore, the velocity of the motorboat after 6s is 24 m/s.
