Answer:
a) k
= 3.6 N/m ; b) 9.43 * 10¹²; c) 1.18 * 10¹¹; d) 75.08 N/m
Step-by-step explanation:
Length of iron bar = L = 2.7m;
side length of cross section = a = 0.07cm = 0.0007m;
x = 2.7 cm = 0.027m;
m = 100kg;
ρ = 7.87gm/cm³;
da = 2.28 * 10⁻¹⁰m;
a)
Fnet = F - mg
where Fnet = 0
So,
F = mg where F=kx
k = mg/x = 100*9.8/0.027
k = 3.6 N/m
b)
Nchain = Aw/Aa =a²/(da)²
= (o.ooo7)²/(2.28 * 10⁻¹⁰)²
= 9.43 * 10¹²
c)
Nbond = L/da = 2.7/2.28 * 10⁻¹⁰
= 1.18 * 10¹¹
d)
Spring stiffness of wire = ksi = (Nbondk)/Nchain
= [(1.18* 10¹¹)(6*10⁴)]/(9.43 * 10¹²)
= 75.08 N/m
Answer:the number of adults that attended the game is 135.
the number of children that attended the game is 31
Step-by-step explanation:
Let x represent the number of adults that attended the game.
Let y represent the number of children that attended the game.
There were 166 paid admissions to a game. This means that
x + y = 166
The price was $2 for adults and $.75 for children. The amount of data taken in was $293.25. This means that
2x + 0.75y = 293.25 - - - - - - - - - - 1
Substituting x = 166 - y into equation 1, it becomes
2(166 - y) + 0.75y = 293.25
332 - 2y + 0.75y = 293.25
- 2y + 0.75y = 293.25 - 332
- 1.25y = - 38.75
y = - 38.75/- 1.25
y = 31
x = 166 - y = x = 166 - 31
x = 135
D is the right answer because g is midpoint of ah and e is midpoint of ab and eg is parallel to bh so eg equals half bh
