Answer:drift velocity is 1.7*10^-4m/s and number of charge is 2.7*10^-24
Explanation:
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE
The velocity of the second glider after the collision is 4.33 m/s rightward.
<h3>
Velocity of the second glider after the collision</h3>
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
where;
- m₁ is mass of first glider
- m₂ is mass of second glider
- u₁ is initial velocity of first glider
- u₂ is initial velocity of second glider
- v is the final velocity of the gliders
(2)(1) + (3)(5) = (2)(2) + 3v₂
17 = 4 + 3v₂
3v₂ = 17 - 4
3v₂ = 13
v₂ = 13/3
v₂ = 4.33 m/s
Thus, the velocity of the second glider after the collision is 4.33 m/s rightward.
Learn more about linear momentum here: brainly.com/question/7538238
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