Suppose the stone is thrown at an angle of 39.0° below the horizontal from the same building as in the Example above. If it strikes the ground 47.8 m away, find the following. (Hint: For part (a), use the equation for the x-displacement to eliminate v0t from the equation for the y-displacement.)(a) the time of flight sThe x coordinate as a function of time is x(t) = vcos(39.0)t, so the initial speed is v0 = Δx/(cos 39.0Δt), where Δx = 47.8 and Δt is the time of flight. Insert this into your equation for y(t) and solve for the time of flight. Note that the answer should be smaller than 3.16227766016838, since the stone is thrown down (and to the right).(b) the initial speed m/s(c) the speed and angle of the velocity vector with respect to the horizontal at impactspeed m/sangle °
No, options are given but I believe the answer would be
In a water cycle Solid state of matter has the particles closest together.
T<u>he direction of motion</u> of the person relative to the water is <u>16.7° north of east.</u>
Why?
We can solve the problem by applying the Pitagorean Theorem, where the first speed (to the north) and the second speed (to the east) corresponds to two legs of the right triangle formed with them. (north and east directions are perpendicular each other)
We can calculate the angle that give the direction using the following formula:

Now, substituting the given information we have:


Hence, we have that <u>the direction of motion</u> of the person relative to the water is 16.7° north of east.
Have a nice day!
Answer:
497.00977 N
3742514.97005
Explanation:
= Density of water = 1000 kg/m³
C = Drag coefficient = 0.09
v = Velocity of dolphin = 7.5 m/s
r = Radius of bottlenose dolphin = 0.5/2 = 0.25 m
A = Area
Drag force

The drag force on the dolphin's nose is 497.00977 N
at 20°C
= Dynamic viscosity = 
Reynold's Number

The Reynolds number is 3742514.97005