Explanation:
We define force as the product of mass and acceleration.
F = ma
It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.
Given Data:
Width of the pool = w = 50 ft
length of the pool = l= 100 ft
Depth of the shallow end = h(s) = 4 ft
Depth of the deep end = h(d) = 10 ft.
weight density = ρg = 62.5 lb/ft
Solution:
a) Force on a shallow end:



b) Force on deep end:



c) Force on one of the sides:
As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.
1) Force on the Rectangular part:




2) Force on the triangular part:

here
h = h(d) - h(s)
h = 10-4
h = 6ft



now add both of these forces,
F = 25000lb + 150000lb
F = 175000lb
d) Force on the bottom:



Answer:Both are correct
Explanation:
Both are correct because
Mechanical efficiency is the dimensionless term which is the ratio of brake horsepower to the Indicated horse Power
Where brake power is the Power obtained at the crankshaft and
Indicated horsepower is the power obtained in the combustion chamber and this power is the loss in the form of friction.
Volumetric efficiency is the ratio of actual fuel intake to the maximum air fuel that could be taken.
Answer:
is the initial velocity of tossing the apple.
the apple should be tossed after 
Explanation:
Given:
- velocity of arrow in projectile,

- angle of projectile from the horizontal,

- distance of the point of tossing up of an apple,

<u>Now the horizontal component of velocity:</u>



<u>The vertical component of the velocity:</u>



<u>Time taken by the projectile to travel the distance of 30 m:</u>



<u>Vertical position of the projectile at this time:</u>



<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>


is the initial velocity of tossing the apple.
<u>Time taken to reach this height:</u>



<u>We observe that </u>
<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>


