Solve for the linear/tangential speed:
<em>a</em> = <em>v</em>²/<em>r</em>
where <em>a</em> = centripetal acceleration, <em>v</em> = speed, and <em>r</em> = radius.
4.7 m/s² = <em>v</em>²/(0.3 m)
<em>v</em>² = (0.3 m) (4.7 m/s²)
<em>v</em> ≈ 3.96 m/s
For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2<em>π</em> (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is
(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s
Take the reciprocal of this to get the period:
1 / (7.46 rev/s) ≈ 0.134 s/rev
So it takes the record about 0.134 seconds to complete one revolution.
Answer:
I = 
(K+5)
Explanation:
Given :
J = k+5
Now selecting a thin ring in the wire of radius "r" and thickness dr.
Current through the thin ring is
dI = J X 2πrdr
dI = (K+5) x 2πrdr
Now integrating we get
I = 
I = (K+5) 2π
I = (K+5) 2π 
I = (K+5)
The law of conservation of mass states that mass can be destroyed during a chemical change is TRUE
Answer:
B) Absorbed energy results in the change in potential energy.
Explanation:
i got it wrong on usatestpreo
U=0
<span>t=10 </span>
<span>a=9.8m/s/s </span>
<span>v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) </span>
<span>you work out the result now</span>
