3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:
(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v
where v is the velocity of the combined players. Solve for v :
450 kg•m/s - 320 kg•m/s = (155 kg) v
v = (130 kg•m/s) / (155 kg)
v ≈ 0.84 m/s
4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that
(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v
where v is the new velocity of the 4-kg ball. Solve for v :
30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v
v = (16.4 kg•m/s) / (4 kg)
v = 4.1 m/s
Answer:
1.6 x 10⁻¹⁹ C
Explanation:
Let us arrange the charges in the ascending order and round them off as follows :-
1.53 x 10⁻¹⁹ C → 1.6x 10⁻¹⁹ C
3.26 x 10⁻¹⁹C → 3.2 x 10⁻¹⁹ C
4.66 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C
5.09 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C
6.39 x 10⁻¹⁹C → 6.4 x 10⁻¹⁹ C
The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.
Here we observe that
2 nd charge is almost twice the first charge
3 rd and 4 th charges are almost 3 times the first charge
5 th charge is almost 4 times the first charge.
This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first charge , 2nd to 5 th charges can be written as 2e, 3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of 1.6 x 10⁻¹⁹ C exists.
B- the acceleration is greater for the more massive rock
