Answer:
Explanation:
Let T be the tension
For linear motion of hoop downwards
mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .
For rotational motion of hoop
Torque by tension
T x R , R is radius of hoop.
Angular acceleration be α,
Linear acceleration a = α R
So TR = I α
= I a / R
a = TR² / I
Putting this value in earlier relation
mg -T = m TR² / I
mg = T ( 1 + m R² / I )
T = mg / ( 1 + m R² / I )
mg / ( 1 + R² / k² )
Tension is less than mg or weight because denominator of the expression is more than 1.
New experiment verify or find previous model
Answer:
Explanation:
Given
Length of rope 
Weight of rope 
weight density
Work done to lift rope 33 m


![W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0](https://tex.z-dn.net/?f=W%3D73.45%5Cleft%20%5B%20%5Cleft%20%28%20%5Cfrac%7Bh%5E2%7D%7B2%7D%5Cright%20%29%5Cright%20%5D%5E%7B33%7D_0)
Answer: Parietal
Explanation: The parietal lobe is where the primary somatosensory cortex is located. This cortex is where all tactile stimulation is processed in the brain and allows to you detect/feel someone scratching your back.
Explanation :
It is given that,
Mass of the car, m = 1000 kg
Force applied by the motor, 
The static and dynamic friction coefficient is, 
Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,




So, the acceleration of the car is
. Hence, this is the required solution.