Answer:
30.63 m
Explanation:
From the question given above, the following data were obtained:
Total time (T) spent by the ball in air = 5 s
Maximum height (h) =.?
Next, we shall determine the time taken to reach the maximum height. This can be obtained as follow:
Total time (T) spent by the ball in air = 5 s
Time (t) taken to reach the maximum height =.?
T = 2t
5 = 2t
Divide both side by 2
t = 5/2
t = 2.5 s
Thus, the time (t) taken to reach the maximum height is 2.5 s
Finally, we shall determine the maximum height reached by the ball as follow:
Time (t) taken to reach the maximum height = 2.5 s
Acceleration due to gravity (g) = 9.8 m/s²
Maximum height (h) =.?
h = ½gt²
h = ½ × 9.8 × 2.5²
h = 4.9 × 6.25
h = 30.625 ≈ 30.63 m
Therefore, the maximum height reached by the cannon ball is 30.63 m
W-APE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative APE. There must be a minus sign in front of APE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.
( The capital A’s in the words are supposed to be triangles ! I also hoped this helped ! Please mark me as brainliest !! )
Answer:
A) ( - 200t + 40 ) volts
B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise
Explanation:
Given data:
magnetic flux (Φm) = 5.0t^2 − 2.0t
number of turns = 20
<u>a) determine induced emf </u>
E = - N 
= - N ( 10t - 2 ) = - 20 ( 10t - 2 )
= - 200t + 40 volts
<u>b) Determine direction of induced current </u>
i) at t = 0
E = - 0 + 40 ( anticlockwise direction )
ii) at t = 0.10
E = -20 + 40 = 20 ( anticlockwise direction )
iii) at t = 1
E = - 200 + 40 = - 160 ( clockwise direction)
iv) at t = 2
E = -400 + 40 = - 360 ( clockwise direction )
Every object will remain at rest
