Answer:
Vo = 4.5 [m/s]
Explanation:
In order to solve this problem, we must use the following equation of kinematics.
where:
Vf = final velocity = 12 [m/s]
Vo = initial velocity [m/s]
a = acceleration = 1.5 [m/s²]
t = time = 5 [s]
Now replacing:
![12=v_{o}+1.5*5\\v_{o}=12- (7.5)\\v_{o}= 4.5[m/s]](https://tex.z-dn.net/?f=12%3Dv_%7Bo%7D%2B1.5%2A5%5C%5Cv_%7Bo%7D%3D12-%20%287.5%29%5C%5Cv_%7Bo%7D%3D%204.5%5Bm%2Fs%5D)
Answer:
Explanation:
Let the tension in the cord be T₁ and T₂ .
for motion of block placed on horizontal table
T₁ = m a , a is acceleration of the whole system .
for motion of hanging bucket of mass m
mg - T₂ = ma
adding the two equation
mg + T₁- T₂ = 2ma
for rotational motion of the pulley
torque = moment of inertia x angular acceleration
(T₂ - T₁) R = I x α , I is moment of inertia of pulley , α is angular acceleration .
(mg - 2ma ) R = I x α
(mg - 2ma ) R = I x a / R
(mg - 2ma ) R² = I x a
mgR² = 2ma R² + I x a
a = mgR² / (2m R² + I )
Since body moves by distance d in time T
d = 1/2 a T²
a = 2d / T²
mgR² / (2m R² + I ) = 2d / T²
mgR²T² = 2d x (2m R² + I )
mgR²T² - 4dm R² = 2dI
m R² ( gT² - 4d ) = 2dI
I = m R² ( gT² - 4d ) ] / 2d .
Myofibrils are composed of long proteins such as actin, myosin, and titin, and other proteins that hold them together. These proteins are organized into thin filaments and thick filaments, which repeat along the length of the myofibril in sections called sarcomeres. Muscles contract by sliding the thin (actin) and thick (myosin) filaments along each other.
