32f. That's because the force is directly proportional to the product of the masses and inversely proportional to the square of the distance. So you get 2•(1/1/4)^2=2•16=32
Considering north as positive Y axis
First 10 miles north means a positive 10 displacement along y direction
Second 20 miles south means a negative 20 displacement along y direction
Third 5 miles south means a positive 5 displacement along y direction
So total displacement = 10 - 20 + 5 = -5 miles displacement
So position vector of the final position = - 5 j
Answer:
30 m/s
Explanation:
Vf = ?
Vi = 0 m/s
a = 20 m/s^2
t = 1.5 s
Plug those values into the equation: Vf = Vi + at
Vf = 0 + (20)(1.5)
Vf = 30 m/s
Answer:
These are Diffraction Grating Questions.
Q1. To determine the width of the slit in micrometers (μm), we will need to use the expression for distance along the screen from the center maximum to the nth minimum on one side:
Given as
y = nDλ/w Eqn 1
where
w = width of slit
D = distance to screen
λ = wavelength of light
n = order number
Making x the subject of the formula gives,
w = nDλ/y
Given
y = 0.0149 m
D = 0.555 m
λ = 588 x 10-9 m
and n = 3
w = 6.6x10⁻⁵m
Hence, the width of the slit w, in micrometers (μm) = 66μm
Q2. To determine the linear distance Δx, between the ninth order maximum and the fifth order maximum on the screen
i.e we have to find the difference between distance along the screen (y₉-y₅) = Δx
Recall Eqn 1, y = nDλ/w
given, D = 27cm = 0.27m
λ = 632 x 10-9 m
w = 0.1mm = 1.0x10⁻⁴m
For the 9th order, n = 9,
y₉ = 9 x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.015m
Similarly, for n = 5,
y₅ = 5x 0.27 x 632 x 10-9/ 1.0x10⁻⁴m = 0.0085m
Recall, Δx = (y₉-y₅) = 0.015 - 0.0085 = 0.0065m
Hence, the linear distance Δx between the ninth order maximum and the fifth order maximum on the screen = 6.5mm
