Answer:
q = 0.036 C
Explanation:
Given that,
Current passes through a defibrillator, I = 18 A
Time, t = 2 ms
We need to find the charge moved during this time. We know that,
Electric current = charge/time
Put all the values,
So, 0.036 C of charge moves during this time.
Explanation:
x = 3.00 – 2.00t + 3.00,
Distance of object at 2 second,
x (t=2) = 3(4) - 2(2) +3
x (t=2) = 12-4 +3
x (t=2) = 11 m
Distance of object at 3 second,
x (t=3) = 3(9) - 2(3) +3
x (t=2) = 27 - 6 + 3
x (t=2) = 24 m
a) the average speed between t = 2.00 s and t = 3.00 s,
Average speed =
Average speed =
Average speed =
Average speed =
Average speed = 11.66
b) the instantaneous speed at t = 2.00 s and t = 3.00 s,
Instantaneous speed =
Instantaneous speed(v) = 6t - 2
Instantaneous speed,v(t=2 to t=3) = 18-2-12+2
Instantaneous speed, v = 6
c) the average acceleration between t = 2.00 s and t = 3.00 s
average acceleration =
average acceleration =
average acceleration = 11.66
d) the instantaneous acceleration at t = 2.00 s and t = 3.00 s
instantaneous acceleration =
instantaneous acceleration =6
instantaneous acceleration = 6
e) for x =0
0 = 3.00 – 2.00t + 3.00
a = 3, b=-2, c=3
t=
t=
t=
general solution of this equation gives imaginary value. Hence, the given object is not at rest.