Answer:
λ = 3 10⁻⁷ m, UV laser
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
let's use trigonometry
tan θ = y / L
as in this phenomenon the angles are small
tan θ =
= sin θ
sin θ = y / L
we substitute
a y / L = m λ
let's apply this equation to the initial data
a 0.04 / L = 1 600 10⁻⁹
a / L = 1.5 10⁻⁵
now they tell us that we change the laser and we have y = 0.04 m for m = 2
a 0.04 / L = 2 λ
a / L = 50 λ
we solve the two expression is
1.5 10⁻⁵ = 50 λ
λ = 1.5 10⁻⁵ / 50
λ = 3 10⁻⁷ m
UV laser
The answer is A 0 degrees is the freezing point of water
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Answer:
The amount of work the factory worker must to stop the rolling ramp is 294 joules
Explanation:
The object rolling down the frictionless ramp has the following parameters;
The mass of the object = 10 kg
The height from which the object is rolled = 3 meters
The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp
Where;
The potential energy P.E. = m × g × h
m = The mass of the ramp = 10 kg
g = The acceleration due to gravity = 9.8 m/s²
h = The height from which the object rolls down = 3 m
Therefore, we have;
P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules
The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules
Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.


Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W
We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)



Part A ) Replacing our values,


Part B ) To find the torque we apply the equation as follow,


