Answer:
The displacement is 386.16m
Explanation:
A seal dives to a depth of 248m. To find displacement, we must calculate the resultant vectors which will give us the displacement
R= sqrt(vector1+vector2)
Since this is a right angle triangle
R= sqrt(248^2 + 296^2)
R= sqrt(149120)
R= 386.16m
Displacement = 386.16m
Newton’s Law: F = MA
A = F/M (change equation)
12.6 N/ 2.4 kg = 5.25
Answer: acceleration is 5.25 m/s^2
Answer:
330.5 m
Explanation:
In this case, the object is launched horizontally at 30° with an initial velocity of 40 m/s .
The maximum height will be calculated as;

where ∝ is the angle of launch = 30°
vi= initial launch velocity = 40 m/s
g= 10 m/s²
h= 40²*sin²40° / 2*10
h={1600*0.4132 }/ 20
h= 661.1/2 = 330.5 m
Answer:

Explanation:
If we express all of the cordinates in their rectangular form we get:
A = (1404.77 , 655.06) m


Since we need C to be (0,0) we stablish that:

That way we make an equation system from both X and Y coordinates:


Replacing values:
With this system we can solve for both Db and Dc and get the answers to the question:


Answer:
a) 91 m/s
b) 111 m/s
Explanation:
v = u + at
v = 128sin60 + (-9.8)(2.0) = 91.25125... m/s
v = √(vx² + vy²) = √((128cos60)² + 91.25125²) = 111.4575... m/s