The voltage and current at the terminals of the circuit element in fig 1.5 are zero for t< 0. for 1.6 0, they are u=80,000te-sor v, 120; -500 -1ste- a

okay i have so many questions. but the big one here is: um wt..f is this mess? ?

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

the answer you are looking for is d. there are no holes for blood to enter or leave the system. your welcome