How many top quark lifetimes have there been in the history of the universe (i.e., what is the age of the universe divided by th

According to Newton's second law, Force equals the rate of change of momentum over time. Momentum change is equal to Force times time. So, F=ma can be rearranged to a=F/m, a more recognizable formulation of Newton's second law
Using a relevant kinematic equation for mass m: V=u+at; where initial speed u=0; thus, acceleration a=F/m gives V=(F/m)xt, which translates to t=mV/F. For mass 2m, applying the same formula: V=u+at; u=0; a=F/2m indicates V=(F/2m)xt, leading to t=2mV/F (possibly double the initial time)
I might have erred somewhere along the line, but the fundamental concept seems valid... using another kinematic equation for m: s=ut + (1/2)at²; with s=d; and initial speed u=0; a=F/m; t=1; results in d=(1/2)(F/m) = F/2m. Similarly, for 2m: s=ut + (1/2)at²; s=d; u=0; a=F/2m; and t=1 gives d=(1/2)(F/2m)=F/4m (half the distance perhaps???? WHAT???!)

Response:

v = 4.08 m/s²

Clarification:

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)

Response:

0.60 m/s

Details:

The average speed between times t = a and t = b can be expressed as:

v_avg = (x(b) − x(a)) / (b − a)

Given the function x(t) = 0.36t² − 1.20t, and considering the interval from 1.0 to 4.0:

v_avg = (x(4.0) − x(1.0)) / (4.0 − 1.0)

v_avg = [(0.36(4.0)² − 1.20(4.0)) − (0.36(1.0)² − 1.20(1.0))] / 3.0

v_avg = [(5.76 − 4.8) − (0.36 − 1.20)] / 3.0

v_avg = [0.96 − (-0.84)] / 3.0

v_avg = 0.60

The average speed calculated is 0.60 m/s.