An electron is negatively charged.
Decreases the input force
Answer:
= 5/9
Explanation:
This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.
B = ρ_liquid g V_liquid
let's write the translational equilibrium condition
B - W = 0
let's use the definition of density
ρ_body = m / V_body
m = ρ_body V_body
W = ρ_body V_body g
we substitute
ρ_liquid g V_liquid = ρ_body g V_body
In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar
V = A h_bogy
Thus
we substitute
5/9 =
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.
When going round a corner your direction changes which means your velocity changes which means there is an acceleration.
