M/s^2 since acceleration is velocity/time=m/s/s=m/s^2
The compound is an indicator
Answer:
The velocity of the ball before it hits the ground = 21.4m/s
Explanation:
The question can be solved as a projectile problem. One of the main characteristics of the parabolic motion is the symmetry it has as long as the initial height is equal to the final height with respect to a reference system, the vertical displacement of the horizontal displacement being independent.
The vertical component of velocity is 13m/s
The horizontal component is 17m/s
The ball's pathway is defined by a right angle triangle .
Using Pythagorean
Let vertical component be a
Let horizontal component be b
Let the speed of the ball before hitting the ground be c
Pythagorean is given by:
a^2 + b^2 = c^2
13^2 + 17^2 = c^2
169 + 289 = c^2
458 = c^2
Sqrt(458) = c
21.4 m/s = c
Answer:
Answer is circular motion. Mark my answer brainliest
Static frictional force = ƒs = (Cs) • (Fɴ)
2.26 = (Cs) • m • g
2.26 = (Cs) • (1.85) • (9.8)
Cs = 0.125
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
1.49 = (Cκ) • m • g
1.49 = (Cκ) • (1.85) • (9.8)
Cκ = 0.0822