65-26
A = 200(1+(0.0275/12)^0.0275(12 x 41 ) graph this in y =
30 M&Ms
1) Gathering the data
12-ounce bag of trail mix = 18 M&Ms
20-ounce bag=?
2) Assuming that the M&Ms are placed in following a constant rate we can write the following proportion:
12 --------------- 18
20 --------------- y
<em>Cross multiplying it due to the first property of proportions</em>
<em />
12y =20 x 18
12y = 360
y=30
3) So, I'd expect to find 30 M&Ms at a 20-ounce bag of trail mix.
Answer:
P_max = 9.032 KN
Step-by-step explanation:
Given:
- Bar width and each side of bracket w = 70 mm
- Bar thickness and each side of bracket t = 20 mm
- Pin diameter d = 10 mm
- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa
- Average allowable shear stress of pin S = 115 MPa
Find:
The maximum force P that the structure can support.
Solution:
- Bearing Stress in bar:
T = P / A
P = T*A
P = (120) * (0.07*0.02)
P = 168 KN
- Shear stress in pin:
S = P / A
P = S*A
P = (115)*pi*(0.01)^2 / 4
P = 9.032 KN
- Bearing Stress in each bracket:
T = P / 2*A
P = T*A*2
P = 2*(120) * (0.07*0.02)
P = 336 KN
- The maximum force P that this structure can support:
P_max = min (168 , 9.032 , 336)
P_max = 9.032 KN
