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=k
x
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:
-36 + 27j
Step-by-step explanation:
−9(4 − 3j)
Distribute
-9 * 4 -9 * (-3j)
-36 + 27j
Answer:
20.5 and 21.5
Step-by-step explanation:
It says 2 significant figures so the least it can possibly be is 0.5 less and the most it can be is 0.5 more as if it was, lets say 20.4 it would round down to 20 not 21.
Ya gotta use the Pythagorean theorem :)
a² + b² = c² (reminder: c is the hypotenuse)
4² + 8² = c²
16 + 64 = 80
√80 = 8.94
H = 8.94
I would say no thats my anwser