Answer:
2.06 m/s
Explanation:
From the law of conservation of linear momentum, the sum of momentum before and after collision are equal. Considering this case where we have frictionless surface, no momentum is lost in the process.
Momentum before collision
Momentum is given by p=mv where m and v represent mass. The initial sum of momentum will be 9v+(27*0.5)=9v+13.5
Momentum after collision
The momentum after collision will be given by (9+27)*0.9=32.4
Relating the two then 9v+13.5=32.4
9v=18.5
V=2.055555555555555555555555555555555555555 m/s
Rounded off, v is approximately 2.06 m/s
Answer:
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Explanation:
At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.
Answer:
Explanation:
Given that,
Mass of the thin hoop
M = 2kg
Radius of the hoop
R = 0.6m
Moment of inertial of a hoop is
I = MR²
I = 2 × 0.6²
I = 0.72 kgm²
Period of a physical pendulum of small amplitude is given by
T = 2π √(I / Mgd)
Where,
T is the period in seconds
I is the moment of inertia in kgm²
I = 0.72 kgm²
M is the mass of the hoop
M = 2kg
g is the acceleration due to gravity
g = 9.8m/s²
d is the distance from rotational axis to center of of gravity
Therefore, d = r = 0.6m
Then, applying the formula
T = 2π √ (I / MgR)
T = 2π √ (0.72 / (2 × 9.8× 0.6)
T = 2π √ ( 0.72 / 11.76)
T = 2π √0.06122
T = 2π × 0.2474
T = 1.5547 seconds
T ≈ 1.55 seconds to 2d•p
Then, the period of oscillation is 1.55seconds
