In this case, the movement is uniformly delayed (the final
rapidity is less than the initial rapidity), therefore, the value of the
acceleration will be negative.
1. The following equation is used:
a = (Vf-Vo)/ t
a: acceleration (m/s2)
Vf: final rapidity (m/s)
Vo: initial rapidity (m/s)
t: time (s)
2. Substituting the values in the equation:
a = (5 m/s- 27 m/s)/6.87 s
3. The car's acceleration is:
a= -3.20 m/ s<span>^2</span>
If it is completely elastic, you can calculate the velocity of the second ball from the kinetic energy
<span>v1 = velocity of #1 </span>
<span>v1' = velocity of #1 after collision </span>
<span>v2' = velocity of #2 after collision. </span>
<span>kinetic energy: v1^2 = v1' ^2 + v2' ^2 (1/2 and m cancel out) </span>
<span>5^2 = 4.35^2 + v2' ^2 </span>
<span>v2 = 2.46 m/s <--- ANSWER</span>
Answer:
yes
Explanation:
because it is independent of eachother
Answer:
E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}
R is radius = Diameter/2 = 0.210m.
At z = 0.2m,
Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,
E can be calculated at distance 0.2m away from the centre of the disk.
Put z = 0.3m and all other values in the equation,
E can be calculated at distance 0.3m away from the centre of the disk
