Water is flowing in a trapezoidal channel at a rate of q = 20 m3/s. the critical depth y for such a channel must satisfy the equation 0=1-(q^2 b)/(ga_c^3 ) where g is gravity, ac is the cross-sectional area, and b is the width of the channel at the surface. for this case, the width and the cross-sectional area can be related to depth y by b = 3 + y and ac = 3y + 1/2y2. use initial guesses for the location of the root of a=0.5 and b= 2.5. the error on the solution should not exceed 1%.

see earlier answer

d₁ = displacement of tern from arctic to antarctic = 11000 miles

d₂ = displacement of tern from antarctic to arctic =   - 11000 miles

d = total displacement for the journey

total displacement for the journey is given as

d = d₁ + d₂

d = 11000 + (-11000)

d = 0 miles

t₁ = time of travel from arctic to antarctic = 3 months = 3 x 30 days = 90 days

t₂ = time of travel from antarctic to arctic = 3 months = 3 x 30 days = 90 days

t = total time of travel

total time of travel is given as

t = t₁ + t₂

t = 90 + 90

t = 180 days

average velocity is given as

average velocity = total displacement/ total time

v = d/t

v = 0/180

v = 0 miles/day

pretty sure it's false

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft