If I knew the answer I would help but I don’t know sorry
Answer:
The stitches and dimples around a baseball and a golf ball respectively, disturbs the air drag on the balls once they are in motion, allowing the them to travel more easily.
Explanation:
The stitches on a baseball disturbs the air drag on the ball when the ball is in motion, allowing the ball to travel more easily. Depending on the orientation of the ball in flight, the drag changes as the flow is disturbed by the stitches.
A smooth ball with no stitches or dimples has more air drag that opposes the motion.
A golf ball is smooth ball with dimples to create a thin turbulent boundary layer of air that clings to the ball's surface. This allows the smoothly flowing air to follow the ball's surface a little farther around the back side of the ball, thereby decreasing the size of the wake, and allowing the ball to travel more easily.
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)
Wood frogs have this adaptation where they accumulate urea in their bodies and convert their liver glycogen to glucose to act as cryoprotectants. This prevents the formation of ice crystals in their bodies that could cause damage cells during freezing in winter.
Answer:
P = 5.22 Kg.m/s
Explanation:
given,
mass of the projectile = 1.8 Kg
speed of the target = 4.8 m/s
angle of deflection = 60°
Speed after collision = 2.9 m/s
magnitude of momentum after collision = ?
initial momentum of the body = m x v
= 1.8 x 4.8 = 8.64 kg.m/s
final momentum after collision
momentum along x-direction
P_x = m v cos θ
P_x = 1.8 x 2.9 x cos 60°
P_x = 2.61 kg.m/s
momentum along y-direction
P_y = m v sin θ
P_y = 1.8 x 2.9 x sin 60°
P_y = 4.52 kg.m/s
net momentum of the body


P = 5.22 Kg.m/s
momentum magnitude after collision is equal to P = 5.22 Kg.m/s