Answer" The average force on the wall be equal to 0.5 N.
Explanation:
To calculate the average force on the wall, we use the following equation:
where,
F = average force = ?N
m = mass of the ball = 0.05kg
= Initial velocity of the ball = -10m/s (negative sign because the ball rebounds)
= Final velocity of the ball = 10m/s
t = time taken by the ball = 2s
Putting values in above equation, we get:
![F=\frac{0.05kg[10-(-10)]m/s}{2s}\\\\F=\frac{0.05kg(20m/s)}{2s}\\\\F=0.5kgm/s^2=0.5N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B0.05kg%5B10-%28-10%29%5Dm%2Fs%7D%7B2s%7D%5C%5C%5C%5CF%3D%5Cfrac%7B0.05kg%2820m%2Fs%29%7D%7B2s%7D%5C%5C%5C%5CF%3D0.5kgm%2Fs%5E2%3D0.5N)
Hence, the average force on the wall will be equal to 0.5N.
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.
Answer:C
Explanation: I studied, and C is correct
