Answer:
Explanation:
Let us study the downward movement of cylinder which accelerates as well as rotates .
A)
If v be the linear downward velocity of cm of cylinder and ω be angular velocity of cylinder
v = ωr , when there is no slippage of string around cylinder.
B &C )
Total kinetic energy = Rotational + linear
= 1/2 Iω² + 1/2 m v²
1/2 x1/2 mr²ω² +1/2 m v²
= 1/4 mv² +1/2 m v²
= 3/4 m v²
For downward acceleration ,
mg - T = ma where T is tension in string.
Rotational movement
Torque = T x r
Tr = I α , I is moment of inertia and α is angular acceleration .
= I a/r
T = I a / r² , Putting this value of T in earlier equation
mg - I a / r² = ma
a (I / r² +m )= mg
a = mg / (I / r² +m )
For cylinders
I = .5 mr²
a = g / (.5 +1)
= g / 1.5
I just finished this question on my test and got it right. 85 is your answer.
Answer:
The cross-sectional area is 
Explanation:
From the question we are told that
The dimension of the cell is 6.9 × 10-4 cm by 1.1 × 10-3 cm = 
Generally the cross-sectional area is mathematically represented as
=>
W = Fd
W = 1225 N x 10 m = 12250
