Answer:
because energy will be lost due to friction, sound, and heat (arguably similar to friction) and ENERGY MUST STAY THE SAME so it is IMPOSSIBLE for the ball to bounce higher than when dropped!
The fourth dimension is technically time. the fourth dimension that you are talking about is actually impossible to comprehend.
Answer:
<h2>C. <u>
0.55 m/s towards the right</u></h2>
Explanation:
Using the conservation of law of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.
Momentum = Mass (M) * Velocity(V)
BEFORE COLLISION
Momentum of 0.25kg body moving at 1.0m/s = 0.25*1 = 0.25kgm/s
Momentum of 0.15kg body moving at 0.0m/s(body at rest) = 0kgm/s
AFTER COLLISION
Momentum of 0.25kg body moving at x m/s = 0.25* x= 0.25x kgm/s
<u>x is the final velocity of the 0.25kg ball</u>
Momentum of 0.15kg body moving at 0.75m/s(body at rest) =
0.15 * 0.75kgm/s = 0.1125 kgm/s
Using the law of conservation of momentum;
0.25+0 = 0.25x + 0.1125
0.25x = 0.25-0.1125
0.25x = 0.1375
x = 0.1375/0.25
x = 0.55m/s
Since the 0.15 kg ball moves off to the right after collision, the 0.25 kg ball will move at <u>0.55 m/s towards the right</u>
<u></u>
The answer to this question is force
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
