Answer:
ΔL = MmRgt / (2m + M)
Explanation:
The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.
ΔL = L − L₀
ΔL = Iω − 0
ΔL = ½ MR²ω
To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.
For the block, two forces act: the weight force mg downward and tension force T upward.
For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.
For the sum of forces in the -y direction on the block:
∑F = ma
mg − T = ma
T = mg − ma
For the sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a/R)
T = ½ Ma
Substituting gives:
mg − ma = ½ Ma
2mg − 2ma = Ma
2mg = (2m + M) a
a = 2mg / (2m + M)
The angular acceleration of the pulley is:
αR = 2mg / (2m + M)
α = 2mg / (R (2m + M))
Finally, the angular velocity after time t is:
ω = αt + ω₀
ω = 2mg / (R (2m + M)) t + 0
ω = 2mgt / (R (2m + M))
Substituting into the previous equations gives:
ΔL = ½ MR² × 2mgt / (R (2m + M))
ΔL = MmRgt / (2m + M)
This involves circuit analysis through simplification of the resistors and capacitors. We need to determine the time constant for each circuit in figures A, B, C, D, and E. This leads to ranking the duration the bulbs remain lit from longest to shortest based on each circuit's time constant. The ranking for the time constants is C > A = E > B > D. Capacitance plays a pivotal role in electrostatics, and devices called capacitors are vital components in electronic circuits. When more charge is applied to a conductor, the voltage escalates proportionately. The capacitance of a conductor is quantified as C = q/v. Adding resistors in series raises resistance while parallel configurations reduce it, conversely increasing capacitance in parallel and diminishing it in series. Thus, circuits with greater time constants take longer to discharge.
The required lift force is approximately 866.92 N. To determine this, we first establish the shark's mass at 92 kg and its density at 1040 kg/m³. The volume of the shark is calculated by dividing mass by density, yielding 0.08846 m³. The buoyant force acting on the shark is then determined by multiplying the volume by the density of water and gravity, resulting in a lift force of 866.92 N.
Response:
The ball remained airborne for 3.896 seconds
Explanation:
Given that
g = 9.8 m/s², representing gravitational acceleration,
If the angle of launch is 45°, the horizontal range will be maximized.
Both horizontal and vertical launch velocities are equal, each equating to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The duration to reach maximum height is half of the flight time.
v = u + at ∵ v = 0 (at maximum height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The total time in the air equals twice the time to reach maximum height
2 t₁ = 3.896 s
The horizontal distance covered is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in the air for 3.896 seconds
