a) Let's call x the direction parallel to the river and y the direction perpendicular to the river.
Dave's velocity of 4.0 m/s corresponds to the velocity along y (across the river), while 6.0 m/s corresponds to the velocity of the boat along x. Therefore, the drection of Dave's boat is given by:

relative to the direction of the river.
b) The distance Dave has to travel it S=360 m, along the y direction. Since the velocity along y is constant (4.0 m/s), this is a uniform motion, so the time taken to cross the river is given by

c) The boat takes 90 s in total to cross the river. The displacement along the y-direction, during this time, is 360 m. The displacement along the x-direction is

so, Dave's landing point is 540 m downstream.
d) If there were no current, Dave would still take 90 seconds to cross the river, because its velocity on the y-axis (4.0 m/s) does not change, so the problem would be solved exactly as done at point b).
Answer:a lid
Explanation bob pulled the lid off the jar of pickles
Answer: y(t)= 1/π^2 sin(6*π^2*t)
Explanation: In order to solve this problem we have to consider the general expression for a harmonic movement given by:
y(t)= A*sin (ω*t +φo) where ω is the angular frequency. A is the amplitude.
The data are: ν= 3π; y(t=0)=0 and y'(0)=6.
Firstly we know that 2πν=ω then ω=6*π^2
Then, we have y(0)=0=A*sin (6*π^2*0+φo)= A sin (φo)=0 then φo=0
Besides y'(t)=6*π^2*A*cos (6*π^2*t)
y'(0)=6=6*π^2*A*cos (6*π^2*0)
6=6*π^2*A then A= 1/π^2
Finally the equation is:
y(t)= 1/π^2 sin(6*π^2*t)
Bird habitats include various habitat types: from the human-related environment (e.g., building area, park, rural area, farmland, and pond) to the natural-related environment (e.g., forestland, grassland, river, stream, and coastland).