Answer:
a) 1. 1365 × 10⁻⁹N
b) 9.1862 × 10⁻⁴⁶N
c) 8.61 × 10⁻¹¹C/Kg
Explanation:
Ke = 8.99× 109 N.m2
/ C2
, G = 6.67 × 10-11 N. m2
/ kg2
a) F = Ke . q₁.q₂/r²
= (8.99× 109 N.m2
/ C2
) ×(1.60×10⁻¹⁹C)²/(4.50×10⁻¹⁰C)²
= 1. 1365 × 10⁻⁹N
b)
F = Gm₁m₂/r²
= (6.67× 10⁻¹¹)×(1.67×10⁻²⁷)²/(4.50×10⁻¹⁰)²
= 9.1862 × 10⁻⁴⁶N
The electric force is larger by 8.0497 ×10³⁷ times
c)
if Keq₁q₂/r² Gm₁m₂/r²,
with q₁=q₂ = q, and m₁ =m₂ = m
Then q/m =
= 
= 8.61 × 10⁻¹¹C/Kg
Answer:
4.0 m/s
Explanation:
In the first part of the run, the athlete runs a distance of

at a speed of

So, the time he/she takes is

In the second part of the run, the athlete covers an additional distance of

with a speed

So, the time taken in this second part is

So, the total distance covered is
d = 300 m + 300 m = 600 m
And the total time taken
t = 100 s + 50 s = 150 s
Therefore, the average speed for the entire trip is
