Many ways for example they look up related things and study them or they can test it and see what happens
The maximum force constant of the spring Kmax is 2337.9 N/m.
<h3>What is force constant of a spring?</h3>
The force constant or spring constant is defined as the force required to stretch or compress a spring such that the displacement in the spring is 1 meter.
Force constant is denoted by K and its unit is N/m.
Where;
K = spring constant
x = displacement
The work done by the spring is given below as follows:
Work done = Fx/2
Kinetic Energy = mv²/2
Force on an inclined plane = mgsinθ
Total force, F = mgsinθ + frictional force
F = 1390 * sin 22° + 515
F = 1035.7 N
Work done = change in KE
Fx/2 = mv²/2
Fx = mv²
m = 1390/9.81 = 141.692
Solving for x;
x = mv²/F
x = 141.692 * 1.8²/1035.7
x = 0.443 m
The maximum force constant of the spring Kmax = 1035.7/0.443
Kmax = 2337.9 N/m
In conclusion, the maximum force constant of the spring is the ratio of the total force and displacement.
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Note that the complete question is given below:
You are designing a delivery ramp for crates containing exercise equipment. The crates of weight 1490 N will move with speed 2.0 m/s at the top of a ramp that slopes downward at an angle 21.0 ∘. The ramp will exert a 533 N force of kinetic friction on each crate, and the maximum force of static friction also has this value. At the bottom of the ramp, each crate will come to rest after compressing a spring a distance x. Each crate will move a total distance of 8.0 m along the ramp; this distance includes x. Once stopped, a crate must not rebound back up the ramp. Calculate the maximum force constant of the spring Kmax that can be used in order to meet the design criteria
the answer would be DISCHARGE
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Answer:
Recall the Diffraction grating formula for constructive interference of a light
y = nDλ/w Eqn 1
Where;
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10-9 m
and n = 1
y₁ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 1) = 0.994m
Q b. How far (m) from the central maximum (m = 0) is the second-order maximum (m = 2) observed?
w = width of slit = 1/15000in =6.67x10⁻⁵in =
6.67x10⁻⁵ x 0.0254m = 1.69x10⁻⁶m
D = distance to screen
λ = wavelength of light
n = order number = 1
Given
y1 = ? from 1st order max to the central
D = 2.66 m
λ = 633 x 10⁻⁹ m
and n = 2
y₂ = 0.994m
Distance (m) from the central maximum (n = 0) is the first-order maximum (n = 2) =1.99m
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