Answer:
The ratio of the model size is 1 : 2000
Explanation:
Given
Real Diameter = 0.012 um
Scale Diameter = 24 um
Required
Determine the scale ratio
The scale ratio is calculated as follows;
Substitute values for real and scale measurements
Divide the numerator and the denominator by 0012um
Represent as ratio
<em>Hence, the ratio of the model size is 1 : 2000</em>
Answer: find the attached files for the answer
Explanation:
The reflected ray appears to have originated from the focal point. We should actually draw a vector from the focal point through the point where the incident ray hits the mirror but we shorten the vector so that its starting point is on the mirror, without changing its angle.
Please find the attached files for the solution
Using the concepts of energy, rotational Newton's second law and rotational kinematics we can find the kinematic energy of the system formed by the disk and the cylindrical axis
KE = 0.23 J
given parameters
to find
This exercise must be solved in parts:
1st part. Endowment kinetic energy is the energy due to the circular motion of an object and is described by the equation
KE = ½ I w²
Where KE is the kinetic energy, I the moment of inertia and w the angular velocity
The moment of inertia is a magnitude that measures the inertia for rotational movement, it is a scalar quantity, therefore it is additive. In this system it is composed of two bodies, the disk and the cylindrical axis, for which the total moment of inertia it is
I_{ total} = I_{ disk} + I_{ cylinder}
the moments of inertia with respect to an axis passing through the center of mass are tabulated
disk I_{disk} = ½ M R²
cylinder I_{cylinder} = ½ m r²
where M and m are the masses of the disk and cylinder respectively, R and r their radii
I_{total} = ½ (M R² + m r²) = ½ M R² ( )
I_{total} = ½ M R² ( ) =
M R² (1 + 0.005 m)
As the shaft mass is much lighter than the disk mass , the last term is very small, which is why we despise it.
I_{total} = ½ M R²
2nd part. Let's use Newton's second law for endowment motion
τ = I α
α = l
τ = F R
α =
With the rotational kinematics expressions, we assume that the system starts from rest (w₀ = 0)
w = w₀ + α t
where w is the angular velocity, alpha is the angular acceleration and t is the time
w = 0 +
we substitute in the kinetic energy equation
KE = ½ I_{total} ( )²
KE = ½
let's substitute
KE =
KE = F² R² t² / M
let's calculate
KE = 12² 0.15² 1.2² / 20
KE = 0.23 J
With the concepts of energy and rotational kinematics we can find the kinetic energy of the system is
KE = 0.23 j
learn more about rotational kinetic energy here: