Answer:
Fn: magnitude of the net force.
Fn=30.11N , oriented 75.3 ° clockwise from the -x axis
Explanation:
Components on the x-y axes of the 17 N force(F₁)
F₁x=17*cos48°= 11.38N
F₁y=17*sin48° = 12.63 N
Components on the x-y axes of the the second force(F₂)
F₂x= −19.0 N
F₂y= 16.5 N
Components on the x-y axes of the net force (Fn)
Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N
Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N
Magnitude of the net force.



Direction of the net force (β)

β=75.3°
Magnitude and direction of the net force
Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis
In the attached graph we can observe the magnitude and direction of the net force
Answer:
750 nm
Explanation:
= separation of the slits = 1.8 mm = 0.0018 m
λ = wavelength of monochromatic light
= screen distance = 4.8 m
= position of first bright fringe =
= order = 1
Position of first bright fringe is given as


λ = 7.5 x 10⁻⁷ m
λ = 750 nm
Answer:

Explanation:
First, denote our known values;

Mass is impulse divided by change in velocity:

Hence, the mass of the ball is 141.30grams
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