Answer:
Explanation:
The angular momentum of an object is given by:
where
m is the mass of the object
v is its velocity
r is the distance of the object from axis of rotation
Here we have:
m = 350 g = 0.35 kg is the mass of the ball
v = 9.0 m/s is the velocity
r = 3.0 m is the distance of the object from axis of rotation (if we take the ground as the centre of rotation)
Therefore, the angular momentum is:
Answer:
The linear velocity of the object is 8.71 m/s.
Explanation:
Given;
mass of the object, m = 1 kg
radius of the circle, r = 3.3 meters
centripetal force, F = 23 N
Centripetal force is given by;
where;
v is the linear velocity of the object
Therefore, the linear velocity of the object is 8.71 m/s.
Answer: An object is in motion when its distance from another object is changing . A reference point is a place or object used for comparison to determine if something is in motion. An object is in motion if it changes position relative to a reference point. You assume that the reference point is stationary, or not moving
Explanation:
Answer:
37.8 m
Explanation:
At point 0, the ball is at height y₀.
At point 1, the ball is at height 30 m.
At point 2, the ball is at height 0 m.
Given:
y₁ = 30 m
y₂ = 0 m
v₀ = 0 m/s
a = -10 m/s²
t₂ − t₁ = 1.5 s
Find: y₀
Use constant acceleration equation.
y = y₀ + v₀ t + ½ at²
Evaluate at point 1.
y₁ = y₀ + v₀ t₁ + ½ at₁²
30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²
30 = y₀ − 5t₁²
Evaluate at point 2.
y₂ = y₀ + v₀ t₂ + ½ at₂²
0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²
0 = y₀ − 5t₂²
y₀ = 5t₂²
Substitute:
y₀ = 5 (1.5 + t₁)²
y₀ = 5 (2.25 + 3t₁ + t₁²)
y₀ = 11.25 + 15t₁ + 5t₁²
30 = 11.25 + 15t₁ + 5t₁² − 5t₁²
30 = 11.25 + 15t₁
t₁ = 1.25
30 = y₀ − 5t₁²
30 = y₀ − 5(1.25)²
y₀ ≈ 37.8
