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 answer is {negative 3 over 5 is located on the right of 0 and −2 is located on the left of 0} its C the third one . maby brainliest
Step-by-step explanation:
You can use the vertical line test (the blue lines on the drawing) and if the other end or if any part of the line touches the blue line, then it isn’t a function.
Answer:
Answer: The mean increases by 3
Step-by-step explanation:
The original data set is
{50, 76, 78, 79, 79, 80, 81, 82, 82, 83}
The outlier is 50 because it is not near the group of values from 76 to 83 which is where the main cluster is.
The original mean is M = (50+76+78+79+79+80+81+82+82+83)/10 = 77
If we take out the outlier 50, the new mean is N = (76+78+79+79+80+81+82+82+83)/9 = 80
So in summary so far
old mean = M = 77
new mean = N = 80
The difference in values is N-M = 80-77 = 3
So that's why the mean increases by 3
