The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to 
.
Response:
- The ratio of the period to is always approximately<u> 2·π : 1</u>
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<h3>How is the value of the ratio of the period to
calculated?</h3>
Given:
The relationship between the period, <em>T</em>, the spring constant <em>k</em>, and the
mass attached to the spring <em>m</em> is presented as follows;
Therefore, the fraction of of the period to , is given as follows;
2·π ≈ 6.23
Therefore;
Which gives;
- The ratio of the period to is always approximately<u> 2·π : 1</u>
Learn more about the oscillations in spring here:
brainly.com/question/14510622
Given:
k = 100 lb/ft, m = 1 lb / (32.2 ft/s) = 0.03106 slugs
Solution:
F = -kx
mx" = -kx
x" + (k/m)x = 0
characteristic equation:
r^2 + k/m = 0
r = i*sqrt(k/m)
x = Asin(sqrt(k/m)t) + Bcos(sqrt(k/m)t)
ω = sqrt(k/m)
2π/T = sqrt(k/m)
T = 2π*sqrt(m/k)
T = 2π*sqrt(0.03106 slugs / 100 lb/ft)
T = 0.1107 s (period)
x(0) = 1/12 ft = 0.08333 ft
x'(0) = 0
1/12 = Asin(0) + Bcos(0)
B = 1/12 = 0.08333 ft
x' = Aω*cos(ωt) - Bω*sin(ωt)
0 = Aω*cos(0) - (1/12)ω*sin(0)
0 = Aω
A = 0
So B would be the amplitude. Therefore, the equation of motion would be x
= 0.08333*cos[(2π/0.1107)t]
Answer: 4 A
Explanation:
Given
Voltage 
Power 
Power is given by 
Insert the values
The rating of fuse is slightly higher than the normal operating conditions. Therefore, a 4 A fuse should be used here.
Answer: They include elements 57-71, or lanthanides, and 89-103, or actinides. ... Inner transition metals have three incomplete outermost electron shells and are all metals.
The answer is c for that question
