It will act upon a buoyant force on the magnitude of which is equal to weight of the fluid
Answer:
2.6×10⁻³ N
Explanation:
From coulomb's law,
F = kq'q/r²................ Equation 1
Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.
From the question,
since q' = q
Then,
F = kq²/r²..................... Equation 2
Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,
Constant: k = 9×10⁹ Nm²/kg².
Substitute into equation 2
F = 9×10⁹(1.79×10⁻¹¹)²/(3.45×10⁻⁵ )²
F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)
F = (28.8369×10⁻¹³)/(11.9025×10⁻¹⁰)
F = 2.6×10⁻³ N.
Constant acceleration of plane = 3m/s²
a) Speed of the plane after 4s
Acceleration = speed/time
3m/s² = speed/4s
S = 12m/s
The speed of the plane after 4s is 12m/s.
b) Flight point will be termed as the point the plane got initial speed, u, 20m/s
Find speed after 8s, v
a = 3m/s²
from,
a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>
t
3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>
8
24 = v - 20
v = 44m/s
After 8s the plane would've 44m/s speed.
