Answer:
The width of the slit is 0.4 mm (0.00040 m).
Explanation:
From the Young's interference expression, we have;
(λ ÷ d) = (Δy ÷ D)
where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.
Thus,
d = (Dλ) ÷ Δy
D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 × m)
d = (3.30 × 563 × ) ÷ (0.0047)
= 1.8579 × ÷ 0.0047
= 0.0003951 m
d = 0.00040 m
The width of the slit is 0.4 mm (0.00040 m).
Answer:
The horizontal component of the truck's velocity is: 23.70 m/s
The vertical component of the truck's velocity is: 3.13 m/s
Explanation:
You have to apply trigonometric identities for a right triangle (because the ramp can be seen as a right triangle where the speed is the hypotenuse), in order to obtain the components of the velocity vector.
The identities are:
Cosα=
Senα=
Where H is the hypotenuse, α is the angle, CA is the adjacent cathetus and CO is the opposite cathetus
The horizontal component of the truck's velocity is:
Let Vx represent it.
In this case, CA=Vx, H=24 and α=7.5 degrees
Vx=(24)Cos(7.5)
Vx=23.79 m/s
The vertical component of the truck's velocity is:
Let Vy represent it.
In this case, CO=Vy, H=24 and α=7.5 degrees
Vy=(24)Sen(7.5)
Vy=3.13 m/s