Answer:
0.82 mm
Explanation:
The formula for calculation an 
bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength 
= 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength 
= 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m 
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Answer:
The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
Explanation:
Given;
initial temperature of the liquid, t₁ = 76.3 +/- 0.4⁰C
final temperature of the liquid, t₂ = 67.7 +/- 0.3⁰C
The change in temperature of the liquid is calculated as;
Δt = t₂ - t₁
Δt = (67.7 - 76.3) +/- (0.3 - 0.4)
Δt = (-8.6) +/- (-0.1)
Δt = 8.6 +/- 0.1 ⁰C
Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
The formula for acceleration is a = F/m; Where: F = force; m = mass
Given: F = .6n; m = .4kg; a = ?
a = F/m
= .6/.4
= 1.5
Therefore, the acceleration of the plate is 1.5 m/s^2
