Answer:
The speed it reaches the bottom is
Explanation:
Given: 
, 
Using the conservation of energy theorem
, 
![m*g*h=\frac{1}{2}*m*(r*w)^2 +\frac{1}{2}*[\frac{1}{2} *m*r^2]*w^2](https://tex.z-dn.net/?f=m%2Ag%2Ah%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2A%28r%2Aw%29%5E2%20%2B%5Cfrac%7B1%7D%7B2%7D%2A%5B%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Ar%5E2%5D%2Aw%5E2)
Solve to w'
The answer to your question is Metal
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
Answer:W = 1.23×10^-6BTU
Explanation: Work = Surface tension × (A1 - A2)
W= Surface tension × 3.142 ×(D1^2 - D2^2)
Where A1= Initial surface area
A2= final surface area
Given:
D1=0.5 inches , D2= 3 inches
D1= 0.5 × (1ft/12inches)
D1= 0.0417 ft
D2= 3 ×(1ft/12inches)
D2= 0.25ft
Surface tension = 0.005lb ft^-1
W = [(0.25)^2 - (0.0417)^2]
W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)
W = 1.23×10^-6BTU
