Answer:
The time taken for the ball to fly up in the air and back down again is 3.058 seconds.
Explanation:
Since the ball ends up at the same vertical distance ( on the ground) as it was at the start of its motion, we can set the total displacement of the ball equal to 0.
Thus, this problem can be simply solved by the following equation of motion:

Here, s = total change in distance = 0 m
u = initial speed = 15 m/s
a = acceleration due to gravity = -9.81 m/s^2
t = time (to be found)
Substituting these values in the equation we get:



t = 3.058 seconds
So, the time taken for the ball to fly up in the air and back down again is 3.058 seconds.
ΔVl = L di/dt
i = i₀e -t/T
di/dt = i₀ × (-1/T) e -t/T
ΔVl = L× (-I/T i₀e -t/T
ΔVl = -L/T i₀e -t/T
b. 15mm, i₀ = 36mA, T = 1.1m
t= Os
ΔVl = 0,491V
C. t = 1ms
ΔVl = 0.198V
t = 2ms
ΔVl = 0.08V
E. t = ms
ΔVl = 0.032V
Answer:
B
Explanation: HELP ME PLEASE, WRITE A ONE PAGE ESSAY TO EXPLAIN THE AUTHOR'S PURPOSE IN WRITING HOM SMART ARE ANIMAL
Answer:
8 units
Explanation:
F=(k*q1*q2)/(r^2)
K is a constant, q1 is charge of 1, q2 is charge of 2, r is distance between the two.