Answer:
a) According to Newton's law of gravitation, as the distance between the Moon and the Earth decreases, the gravitational attraction increases and vice versa
The gravitational force of the Moon on the Earth causes the Earth to be slightly bulged on the side directly facing the Moon
The gravitational force also pulls the water bodies on the Earth's surface towards the Moon in the same manner and the effect is more pronounced due to the ability of the liquid water to assume a shape based on the magnitude of the gravitational field attracting it
Therefore, the region where the Moon is closest to the Earth we have a high tide as the water level rises and the region which is perpendicular to where the Moon is located has a low tide
b) The two special types of tides are
1) The neap tide
2) The spring tide
Neap tide
Neap tide occurs when the Sun and Moon are 90° apart from each other when they are viewed by an observer from Earth
The gravitational pull of the Sun cancels (partially) the effect of the gravitational pull and tidal force of the Moon, resulting in minimum tidal range
Spring Tide
Spring tide occurs when the Earth, the Moon, and the Sun are simultaneously inline, such that the Sun reinforces the gravitational pull and tidal force of the Moon, resulting in a maximum tidal range
Explanation:
Answer:
Explanation:
We have here values from SI and English Units. I will convert the units to English Units.
We hace for the power P,
we have other values such 
and 
(specific weight of the water), and 0.85 for \eta
We need to figure the flow rate of the water (V) out, that is,
Where 
is the turbine efficiency, at which is,
Replacing,
With this value (the target of this question) we can also calculate the mass flow rate of the waters,
through the density and the flow rate,
converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,
To reach a vertical height of 13.8 ft against gravity, which has an acceleration of 32 ft/s^2, the required vertical speed can be calculated from the equation:
vi^2 - vf^2 = 2*g*h
Given that it has vf = 0 (it is not moving vertically at its maximum height), g = 32, and h = 13.8, we can solve for vi:
vi^2 = 29.72 ft/s
This is only its vertical speed, so this is equivalent to its original speed multiplied by the sine of the angle:
29.72 ft/s = (v_original)*(sin 42.2<span>°</span>)
v_original = 44.24 ft/s
Converting to m/s, this can be divided by 3.28 to get 13.49 m/s.
Answer:
Hoop will reach the maximum height
Explanation:
let the mass and radius of solid ball, solid disk and hoop be m and r (all have same radius and mass)
They all are rolled with similar initial speed v
by the law of conservation of energy we can write
for solid ball
![[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{ball}\omega^2= mgh_{ball}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%2B%5Cfrac%7B1%7D%7B2%7DI_%7Bball%7D%5Comega%5E2%3D%20mgh_%7Bball%7D)
putting 
in the above equation and solving we get
now for solid disk
![[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{disk}\omega^2= mgh_{disk}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%2B%5Cfrac%7B1%7D%7B2%7DI_%7Bdisk%7D%5Comega%5E2%3D%20mgh_%7Bdisk%7D)
putting 
in the above equation and solving we get
for hoop
![[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{hoop}\omega^2= mgh_{hoop}](https://tex.z-dn.net/?f=%5Btex%5D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%2B%5Cfrac%7B1%7D%7B2%7DI_%7Bhoop%7D%5Comega%5E2%3D%20mgh_%7Bhoop%7D)
putting 
in the above equation and solving we get
clearly from the above calculation we can say that the Hoop will reach the maximum height
