1) Horizontal velocity, Vx
Vx is constant and equal to 5.0 m/s
The graph Vx(t) = 5.0 m/s is a horizontal line (parallel to the horizontal t-axis). that intercepts the vertical-axis at 5.0 and runs from t = 0 to t = 2.
2) Vertical velocity, Vy
Vy = gt = 10t
d = at^2 / 2 => tmax = √(2d / a) = √(2*20m/10m/s^2) = 2 s
Graph Vy is an inclined line with slope 10 m/s^2, that runs from t =0 to t = 2 s, and passes through the points (0,0), (1,10), and (2,20).
We are given that the wavelength ʎ is from 400 nm to 700
nm. The formula for this is:
d sin a =m * ʎ
where,
d = slit separation = 1 mm / 750 lines = 1/750
a = angle
m = 1
ʎ = 400 nm to 700 nm = 0.0004 mm to 0.0007 mm
Rewriting the formula in terms of angle a:
a = sin^-1 (m ʎ / d)
when ʎ = 0.0004 mm
a = sin^-1 (0.0004 / (1/750))
a = 17.46°
when ʎ = 0.0007 mm
a = sin^-1 (0.0007 / (1/750))
a = 31.67°
Hence the range of angles is from 17.46° to 31.67<span>°.</span>
Answer:OBSERVING IS WHATCHING A OBJECT VERY CLOSELY AND EXPIERIMENTING IS WER YOUTEST ON A CERTAIN THING OR CREATURE OR MASS OR ELEMENT
Explanation:IM THE MYSTERY MAN WHOOSH
The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
Answer:
Explanation:
So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.
So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then
v = 25.8 + (-1.66×8.3)
v =12.022 m/s.
So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.
