This question is incomplete the complete question is
A diver bounces straight up from a diving board, avoiding the diving board on the way down, and falls feet first into a pool. She starts with a velocity of 4.00 m/s and her takeoff point is 1.80 m above the pool. (a) What is her highest point above the board? (b) How long a time are her feet in the air? (c) What is her velocity when her feet hit the water?
Answer:
(a) Xs=0.459m
(b) t=0.984 s
(c) Vc=6.65 m/s
Explanation:
(a) To reach maximum distance
(b) For Time
To find t we must find t1 and t2
as
t=t1+t2
For T1
For T2
For Total Time
t=t1+t2
t=0.306+0.6789
t=0.984s
(c) To find Vc
Vc=Vb+gt2
Vc=(0)+(9.8)(0.6789)
Vc=6.65 m/s
Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity 
will be expressed by the following equation:
Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is:
C. The slope of the ant's displacement vs. time graph.
Light is a form of energy.
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The pressure at 100 meters below the surface of sea water with a density of 1150kg is 145.96 psi.
