Since 1m/s=3.6 km/h, we can conclude that 10.0m/s = 36 km/h
Answer:
The magnitude of the impulse delivered to the baseball is 7.0 Ns
Explanation:
Given;
mass of the foul ball, m = 0.14 kg
initial velocity, u = 40 m/s
final velocity, v = 30 m/s in perpendicular direction
Impulse is given as change in momentum;
initial momentum in horizontal direction, Pi = mu
Pi = 0.14 x 40 = 5.6 Ns
final momentum in perpendicular direction, Pf = mv
Pf = 0.14 x 30
Pf = 4.2 Ns
The resultant impulse is given by;
J² = 5.6² + 4.2²
J² = 49
J = √49
J = 7.0 Ns
Therefore, the magnitude of the impulse delivered to the baseball is 7.0 Ns
Answer:
w = 25.05 rad / s
, α = 0.7807 rad / s²
, θ = 1972.75
Explanation:
This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off
θ = w₀ t - ½ α t²
α = (w₀t - θ) 2/t²
let's reduce the magnitudes to the SI system
w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s
θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad
let's calculate the angular acceleration
α = (55.5 39 - 1570.8) 2/39²
α = 0.7807 rad / s²
having the acceleration we can calculate the final speed
w = w₀ - ∝ t
w = 55.5 - 0.7807 39
w = 25.05 rad / s
the time to stop w = 0
0 = wo - alpha t
t = wo / alpha
t = 55.5 / 0.7807
t = 71.09 s
the angle traveled
w² = w₀⁹ - 2 α θ
w = 0
θ = w₀² / 2α
let's calculate
θ = 55.5 2 / (2 0.7807)
θ = 1972.75
Answer:
So, put simply, reflection is light bouncing off surfaces. The law of reflection determines how the light bounces off a surface. The law of reflection states that the angle of incidence is equal to the angle of reflection.
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<em><u>Hope this helps..</u></em></h2>
