Pressure= hqg
H=depth
q=density
g=gravity
h=0.2
q=7
g=10
0.2*7*10= 14pa
FINAL ANSWER = 14pa
Answer:
t = 123.59s
Explanation:
For the launch pad section:
Vf = Vo + a*t where Vo=0.
Vf = 35*25 = 875m/s
The distance traveled during the launch:

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

where d= 10937.5m; Vo=875m/s.
Solving for t:
t1 = -11.093s t2 = 98.59s
So, the total time of flight will be:

Answer:
a) 2.933 m
b) 4.534 m
Explanation:
We're given the equation
v(t) = -0.4t² + 2t
If we're to find the distance, then we'd have to integrate the velocity, since integration of velocity gives distance, just as differentiation of distance gives velocity.
See attachment for the calculations
The conclusion of the attachment will be
7.467 - 2.933 and that is 4.534 m
Thus, The distance it travels in the second 2 sec is 4.534 m
Answer:
393.6m/s
Explanation:
Given parameters:
Acceleration = 8.5m/s²
Distance = 300m
Final velocity = 400m/s
Unknown:
Initial velocity = ?
Solution:
To solve this problem, we use the expression below;
v² = u² + 2as
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance
So;
v² - 2as = u²
u² = v² - 2as
u² = 400² - (2 x 8.5 x 300)
u = 393.6m/s