Answer:
<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>
</h2>
Explanation:
Part a)
As we know that drag force is given as
so we have
So acceleration of the ball is
Part B)
As per kinematics we know that
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.
Answer:
C: equal to mg
Explanation:
in free-fall, gravity is always the net force on an object
Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (
=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ = 
q₂ =
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀
