Answer:An inelastic collision is one in which the internal kinetic energy changes (it is not conserved). A collision in which the objects stick together is sometimes called perfectly inelastic because it reduces internal kinetic energy more than does any other type of inelastic collision.People sometimes think that objects must stick together in an inelastic collision. However, objects only stick together during a perfectly inelastic collision. Objects may also bounce off each other or explode apart, and the collision is still considered inelastic as long as kinetic energy is not conserved.
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Explanation:
Answer:
36 N/m
Explanation: I got the answer right on the test ight noo cap
Answer:
a) 7200 ft/s²
b) 140 ft
c) 3.7 s
Explanation:
(a) Average acceleration is the change in velocity over change in time.
a_avg = Δv / Δt
We need to find what velocity the puck reached after it was hit by the hockey player.
We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s². Therefore:
v² = v₀² + 2a(x − x₀)
(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)
v₀² = 5200 ft²/s²
v₀ = 20√13 ft/s
So the average acceleration impacted to the puck as it is struck is:
a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)
a_avg = 2000√13 ft/s²
a_avg ≈ 7200 ft/s²
(b) The distance the puck travels before stopping is:
v² = v₀² + 2a(x − x₀)
(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)
x = 140 ft
(c) The time the puck takes to travel 10 ft without friction is:
t = (10 ft) / (20√13 ft/s)
t = (√13)/26 s
The time the puck travels over the rough ice is:
v = at + v₀
(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)
t = √13 s
So the total time is:
t = (√13)/26 s + √13 s
t = (27√13)/26 s
t ≈ 3.7 s
Answer:
Incomplete question
The complete question is
A Ferris wheel is a vertical, circular amusement ride with radius 6.0 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 9.6 s. Consider a rider whose mass is 96 kg.
At the bottom of the ride, what is the rate of change of the rider's momentum?
Explanation:
Radius of wheel is 6m
Rider mass=96kg
He completes one revolution in 9.6s
Let get angular velocity (w)
1 Revolution =2πrad
θ=2πrad
w= θ/t
w=2π/9.6
w=0.654rad/s
Linear speed is give as
v=wr
v=0.654×6
v=3.93m/s
Centripetal acceleration a
a=rw²
a=6×0.654²
a=2.57m/s²
Acceleration due to gravity g=9.81m/s²
According to Newton's second law of motion net force acting on the rider at the bottom of the ride is given by: the two force acting at the bottom is the normal and the weight of the rider
ΣF = ma
N-W=ma
N-mg=ma
N=ma+mg
N=m(a+g)
N=96(2.57+9.81)
N=1188.48 N
Therefore the rate of change of momentum at the bottom of the ride is 1188.48 N.
Answer:
The turning points are those instants, moments or situations that happen in an absolutely unexpected way, as a result of which your life changes ... and nothing is the same as before.
