The initial velocity of the ball is 0. Applying:
v = u + at
v = 0 + 229 x 0.08
v = 18.3 m/s
a)
Vx = Vcos(∅)
Vx = 18.3cos(52.3)
Vx = 11.2 m/s
b)
Vy = Vsin(∅)
Vy = 18.3sin(52.3)
Vy = 14.5 m/s
Answer:
North of west
Explanation:
Given
40,000-ton luxury line traveling 20 knots towards west and
60,000 ton freighter traveling towards North with 10 knots
suppose v is the common velocity after collision
conserving momentum in west direction

suppose the final velocity makes \theta angle with x axis

Conserving Momentum in North direction


divide 1 and 2


so search in the area
North of west to find the ship
C it reduces the amount of useful work done on objects move it up the ramp
Answer:
Fy = 14.3 [N]
Explanation:
To be able to solve this problem we must know that the force is a vector and has magnitude and direction, therefore it can be decomposed into the force in the X & y components:
When we have the components on the horizontal and vertical axes we must use the Pythagorean theorem.

where:
F = 15 [N]
Fx = horizontal component = 4.5 [N]
Fy = vertical component [N]
![15=\sqrt{4.5^{2}+F_{y}^{2}}\\ 15^{2}= (\sqrt{4.5^{2}+F_{y}^{2}})^{2} \\225 = 4.5^{2}+F_{y} ^{2}\\ F_{y}^{2} =225 -4.5^{2}\\ F_{y}^{2}=204.75\\F_{y}=\sqrt{204.75}\\ F_{y}=14.3 [N]](https://tex.z-dn.net/?f=15%3D%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%5C%5C%2015%5E%7B2%7D%3D%20%28%5Csqrt%7B4.5%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%29%5E%7B2%7D%20%5C%5C225%20%3D%204.5%5E%7B2%7D%2BF_%7By%7D%20%5E%7B2%7D%5C%5C%20%20F_%7By%7D%5E%7B2%7D%20%3D225%20-4.5%5E%7B2%7D%5C%5C%20F_%7By%7D%5E%7B2%7D%3D204.75%5C%5CF_%7By%7D%3D%5Csqrt%7B204.75%7D%5C%5C%20%20F_%7By%7D%3D14.3%20%5BN%5D)