Answer:
11:1
Explanation:
At constant acceleration, an object's position is:
y = y₀ + v₀ t + ½ at²
Given y₀ = 0, v₀ = u, and a = -g:
y = u t − ½g t²
After 6 seconds, the ball reaches the maximum height (v = 0).
v = at + v₀
0 = (-g)(6) + u
u = 6g
Substituting:
y = 6g t − ½g t²
The displacement between t=0 and t=1 is:
Δy = [ 6g (1) − ½g (1)² ] − [ 6g (0) − ½g (0)² ]
Δy = 6g − ½g
Δy = 5½g
The displacement between t=6 and t=7 is:
Δy = [ 6g (7) − ½g (7)² ] − [ 6g (6) − ½g (6)² ]
Δy = (42g − 24½g) − (36g − 18g)
Δy = 17½g − 18g
Δy = -½g
So the ratio of the distances traveled is:
(5½g) / (½g)
11 / 1
The ratio is 11:1.
I think it’s D) all of the above
Answer:
The pressure at point 2 is 
Explanation:
From the question we are told that
The speed at point 1 is 
The gauge pressure at point 1 is 
The density of water is 
Let the height at point 1 be 
then the height at point two will be
Let the diameter at point 1 be 
then the diameter at point two will be
Now the continuity equation is mathematically represented as
Here 
are the area at point 1 and 2
Now given that the are is directly proportional to the square of the diameter [i.e 
]
which can represent as
=> 
where c is a constant
so 
=> 
=> 
Now from the continuity equation
=> 
=> 
Generally the Bernoulli equation is mathematically represented as
So
=> 
substituting values
Answer:
A.) 1430 metres
B.) 80 seconds
Explanation:
Given that the train accelerates from rest at 1.1m/s^2 for 20s. The initial velocity U will be:
U = acceleration × time
U = 1.1 × 20 = 22 m/s
It then proceeds at constant speed for 1100 m
Then, time t will be
Time = distance/ velocity
Time = 1100/22
Time = 50 s
before slowing down at 2.2m/s^2 until it stops at the station.
Deceleration = velocity/time
2.2 = 22/t
t = 22/2.2
t = 10s
Using area under the graph, the distance between the two stations will be :
(1/2 × 22 × 20) + 1100 + (1/2 × 22 × 10)
220 + 1100 + 110
1430 m
The time taken between the two stations will be
20 + 50 + 10 = 80 seconds
