Answer:
The answer to the question is
4433.416 kJ
See explanation below
(3-y)²+r² = 3² or
6y-y² = r²
r =√(6y-y²)
The volume of a small section of height Δy = Δy ×(√(6y-y²))²×π
For water with density of 1000 kg/m³, the mass of the slice
= 1000×Δy ×(√(6y-y²))²×π and since force = mass × acceleration we have
1000×Δy ×(√(6y-y²))²×π ×g = 1000×Δy ×(√(6y-y²))²×π ×9.81
The work done to move a unit height of y+1 = Force × Distance
W = 1000×Δy ×(√(6y-y²))²×π ×9.81 × (y+1)
Integrating the entire section of the sphere that is 2×r high, or from 0 to 6 we get
W =
= 
![= 9810*\pi *[\frac{5y^{3} }{3} -\frac{y^{4} }{4} +3y^{2} ]^{6} _{0}](https://tex.z-dn.net/?f=%3D%209810%2A%5Cpi%20%2A%5B%5Cfrac%7B5y%5E%7B3%7D%20%7D%7B3%7D%20-%5Cfrac%7By%5E%7B4%7D%20%7D%7B4%7D%20%2B3y%5E%7B2%7D%20%5D%5E%7B6%7D%20_%7B0%7D)
=9810×π×144 =4433416 J
Answer:
Explanation:
Given
Launch angle =u
Initial Speed is 
Horizontal acceleration is 
At maximum height velocity is zero therefore
Total time of flight 
During this time horizontal range is
For maximum range 
![\frac{\mathrm{d} R}{\mathrm{d} u}=\frac{2v_0^2}{g}\left [ \cos 2u-\frac{a}{g}\sin 2u\right ]=0](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20R%7D%7B%5Cmathrm%7Bd%7D%20u%7D%3D%5Cfrac%7B2v_0%5E2%7D%7Bg%7D%5Cleft%20%5B%20%5Ccos%202u-%5Cfrac%7Ba%7D%7Bg%7D%5Csin%202u%5Cright%20%5D%3D0)
(b)If a =10% g
thus 
Answer:
Explanation:
Width of central diffraction peak is given by the following expression
Width of central diffraction peak= 2 λ D/ d₁
where d₁ is width of slit and D is screen distance and λ is wave length.
Width of other fringes become half , that is each of secondary diffraction fringe is equal to
λ D/ d₁
Width of central interference peak is given by the following expression
Width of each of bright fringe = λ D/ d₂
where d₂ is width of slit and D is screen distance and λ is wave length.
Now given that the central diffraction peak contains 13 interference fringes
so ( 2 λ D/ d₁) / λ D/ d₂ = 13
then ( λ D/ d₁) / λ D/ d₂ = 13 / 2
= 6.5
no of fringes contained within each secondary diffraction peak = 6.5
Answer:
A 10 N force pointing up
Explanation:
If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).
Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).
Then there must also be a 10 N force pointing up acting on the object.
Explanation:
A. Walk the person around
