B would be the correct answer
Answer:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Explanation:
Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:
(1)
Where:
, 
- Initial and final momemtums of the small object, measured in kilogram-meters per second.
, 
- Initial and final momentums of the big object, measured in kilogram-meters per second.
If we know that 
, 
and 
, then the final momentum of the big object is:
The magnitude of the large object's momentum change is:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Answer:
a) I = 13.38 kg m / s, b) F = 1,373 10³ N
Explanation:
The impulse is given by the relation
I = ∫ F dt = Δp
I = p_f -p₀
I = m (v_f - v₀)
take the ball's exit direction as positive, whereby the ball velocities
v₀ = -90mph, the final velocity v_f = + 54 m / s
Let's reduce the units to
I = 0.142 [54- (-40.23) ]
the SI system
v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s
m = 142 g (1kg / 1000) = 0.142 kg
we calculate
I = 0.142 [54- (-40) ]
I = 13.38 kg m / s
b) let's use the definition of momentum
I = ∫ F .dt
I = F ∫ dt
F = I / t
F = 13.38 / 0.008
F = 1,373 10³ N
For a circular aperture, the first minima (n=1) as an angular separation from the peak of the central maxima given by
Sinθ = 1.22λ / d
Where,
d is the aperture or pupil diameter
d = 4.69 mm = 4.69 × 10^-3m
λ is the wavelength
λ = 545 nm = 545 × 10^-9 m
Then,
Sinθ = 1.22λ / d
Sinθ = 1.22 × 545 × 10^-9 / 4.69 × 10^-3
Sinθ = 1.418 × 10^-4 rad
Then, the head light sources have the same angular separation θ from the eye as the image have inside the eye.
For the headlight
Sinθ ≈ light separation / distantce for the eye
Light separation is give as x = 0.659 m
And let the distance of the eye be D
Then,
Sinθ = x / D
Make D subject of formula
D = x / Sinθ
D = 0.695 / 1.418 × 10^-4
D = 4902.316m
To km, 1km = 1000m
D ≈ 4.9 km
