Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft × 4 sorings
K = 100 lb/ft
Amplitude of the microscope ![\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}]](https://tex.z-dn.net/?f=%5Cfrac%7BX%7D%7BY%7D%3D%20%5B%5Cfrac%7B1%2B2%20%5Cepsilon%20%28%5Comega%2F%20W_n%29%5E2%7D%7B%281-%28%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%29%5E2%2B%282%20%5Cepsilon%20%20%5Cfrac%7B%5Comega%7D%7BW_n%7D%29%5E2%7D%5D)
where;


= 
= 4.0124
replacing them into the above equation and making X the subject of the formula:



Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = 
Answer:
The correct answer is B)
Explanation:
When a wheel rotates without sliding, the straight-line distance covered by the wheel's center-of-mass is exactly equal to the rotational distance covered by a point on the edge of the wheel. So given that the distances and times are same, the translational speed of the center of the wheel amounts to or becomes the same as the rotational speed of a point on the edge of the wheel.
The formula for calculating the velocity of a point on the edge of the wheel is given as
= 2π r / T
Where
π is Pi which mathematically is approximately 3.14159
T is period of time
Vr is Velocity of the point on the edge of the wheel
The answer is left in Meters/Seconds so we will work with our information as is given in the question.
Vr = (2 x 3.14159 x 1.94m)/2.26
Vr = 12.1893692/2.26
Vr = 5.39352619469
Which is approximately 5.39
Cheers!
Answer:
Explanation:
1) Hypermetropia (better known as Farsighted- this is why nearby objects seem blurry for him)
2) In such instances, image are typically formed farther from the near point
3) Such defects are quite common so there are common procedures such as using convex lens which can restore the sight to normal.
Answer:
Explanation:
Given that the grand stone has initial angular velocity of
w(ini)= 6rad/
And it has a final angular velocity of
w(fin)=12.20rad/sec
Time taken is t=16s
Using equation of angular motion
To get angular acceleration (α)
w(fin)=w(ini)+αt
12.20=6+16α
16α=12.20-6
16α=6.2
α=6.2/16
α=0.3875rad/sec²
The angular acceleration is 0.39rad/s²
Angle that he turn using
w(fin)²=w(ini)²+2αθ
12.2²=6²+2×0.3875θ
12.2²-6²=0.775θ
0.775θ=112.84
Then, θ=112.84/0.775
θ=145.6radian
The angular displacement is 145.6rad