0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
<h3>Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.</h3>
Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
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Answer: food, water, habitat, and mate.
Explanation: The common limiting factors in an ecosystem are food, water, habitat, and mate. The availability of these factors will affect the carrying capacity of an environment. As population increases, food demand increases as well
Answer:
Explanation:
Maximum force of friction possible = μmg
= .65 x 3.8 x 9.8
= 24.2 N
u = 72 x 1000 / 60 x 60
= 20 m /s
v² = u² - 2as
a = 20 x 20 / (2 x 30)
= 6.67 m / s²
force acting on it
= 3.8 x 6.67
= 25.346 N
Friction force possible is less .
So friction will not be able to prevent its slippage
It will slip off .
Answer:
a)W=8.333lbf.ft
b)W=0.0107 Btu.
Explanation:
<u>Complete question</u>
The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.
Solution
Preload = F₀=0 lbf
Spring constant k= 200 lbf/in
Initial length of spring x₁=0
Final length of spring x₂= 1 in
At any point, the force during deflection of a spring is given by;
F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.
Change to lbf.ft by dividing the value by 12 because 1ft=12 in
100/12 = 8.333 lbf.ft
work required to compress the spring, W=8.333lbf.ft
The work required to compress the spring in Btu will be;
1 Btu= 778 lbf.ft
?= 8.333 lbf.ft----------------cross multiply
(8.333*1)/ 778 =0.0107 Btu.
