Well I don't know. Let's actually LOOK at the picture and see if that helps.
A, B, C, and D all have the same TOTAL length, but A has the most waves crammed into that same total length.
By golly, that means the length of <u><em>each</em></u> wave in A must be shorter than each wave in B, C, or D.
The correct choice is <em> A </em>. Looking at the picture did the trick !
Answer:
"8 units" is the appropriate answer.
Explanation:
According to the question,
Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.
Now,
The total energy will be:
= 
= 
The total number of particles will be:
= 
= 
hence,
Energy of each A particle or each B particle will be:
= 
= 
Answer:
Final mass=0.89kg
Final pressure=5.6bar
Explanation:
To find mass,m=v/v1
But v1=vf + x(vg-vf)
Vf= 0.001093m^3/kg
Vg= 0.3748m^3/kg
V1= 0.001093+0.5(0.3748-0.001093)
V1= 0.225m^3/kg
M= 0.20/0.225 =0.89kg
Final pressure will be:
V/V1= P/P1
Cross multiply
VP1=V1P
P1= 0.225×5/0.2
P1=:5.6 bar
Answer: It's hard to say without characterizing the collision. But it will be either A if the collision is totally in-elastic, or B if the collision is totally elastic. It could be anywhere in between for partially elastic collisions.
Explanation:
momentum is conserved, so initial system momentum will be left to right.
The velocity of the center of mass is 50(5) / 550 = 0.4545... m/s
In an elastic collision, the lead ball will move off at twice that speed or 0.91 m/s to the right.
The steel ball will bounce back and move away at 0.91 - 5 = -4.1 m/s . The negative sign indicates the steel ball has reversed course and has negative momentum
In a totally in-elastic collision, both balls would move to the right at 0.45 m/s. The steel ball will still have positive momentum.