If you pack more mass into a certain volume, the density will increase.
The Atwood's machine is in motion starting from rest, then Vf = Vo + a(t).
<span>Final Velocity is given as 6.7 m/s and the time is 1.9 s thus 6.7= 0+ a(1.9) </span>
<span>then a = 6.7/1.9 = 3.526 m/s². </span>
<span>The Atwood's Machine also has the formula d= distance = 1/2a(t²) </span>
<span>distance given is 6.365 m , then 6.365 = 1/2 a (1.9)², </span>
<span>a = 3.526 m/s² the same acceleration. </span>
<span>a= g(m1-m2) / m1+m2) </span>
<span>m1a + m2a = m1g - m2g </span>
<span>m1a - m1g = -m2g - m2a </span>
<span>3.526 m1 - 9.81 m1 = -9.81m2 - 3.526 m2 </span>
<span>-6.28 m1 = -13.34 m2 </span>
<span>0.47 m1= m2 </span>
<span>if 24J = 1/2mv² </span>
<span>then 24J = 1/2 m1 ( 6.7)² </span>
<span>48/ 44.89 = m1 </span>
<span>1.069 kg = m1 , then </span>
<span>0.47(1.069) = m2 </span>
<span>0.503 kg = m2</span>
Answer:
Alpha = ω^2 R where R is radius of blade
g = w^2 r where r is distance from center
ω^2 R = 11.5 ω^2 r
R / r = 11.5 / 9.8 = 1.17
Or r = .852 R
Since the angular acceleration depends on both R and ω it seems that one can only get r as it depends on R
Frequency=1/T, where T is the time period.
Reciprocal of frequency = T or it can be written as T = 1/f, where f is frequency.
To solve this problem it is necessary to apply the concepts related to heat exchange in the vegetable and water.
By definition the exchange of heat is given by
where,
m = mass
c = specific heat
= Change in temperature
Therefore the total heat exchange is given as
Our values are given as,
Total mass is 
= 200lb ,however the mass of solid vegetable and water is given as,
Replacing at our equation we have,
Therefore the heat removed is 22411.2 Btu
