Answer:
Step-by-step explanation:
Given
Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height
Suppose they have same mass and radius
time Period is given by
,where h=height of release
a=acceleration
Where I=moment of inertia
a for hoop
a for Uniform solid cylinder
a for spherical shell
a for Uniform Solid
time taken will be inversely proportional to the square root of acceleration
thus first one to reach is Solid Sphere
second is Uniform solid cylinder
third is Spherical Shell
Fourth is hoop
If A, B and C are collinear, then
1) if B is between A and C:
AC = AB + BC
AC = 48 + 22 = 70
2) if C is between A and B:
AB = AC + BC
48 = AC + 22 |-22
AC = 26
3) if A is between B and C:
BC = AB + AC
22 = 48 + AC |-48
AC = - 26 < 0 FALSE
Answer:
if B is between A and C, then AC = 70
if C is between A and B, then AC = 26
Answer:
a. Diego
b. Lin
Step-by-step explanation:
a. Diego
b. Lin
It's pretty easy just do some subtraction.
Actually look and pay attention in class this is like grade 1 math.
Answer:
Since the balls are drawn with replacement, it means the individual probability remain constant,
I have solved this problem on paper (Figures Attached).
Thanks.
