Answer:
F = 37.8 × 10^(6) N
Explanation:
The charges are 0.06 C and 0.07 C.
Thus;
Charge 1; q1 = 0.06 C
Charge 2; q2 = 0.07 C
Distance between them; r = 3 m
Formula for the force in between them is;
F = kq1•q2/r²
Where k is a constant = 9 × 10^(9) N.m²/C²
Thus;
F = (9 × 10^(9) × 0.06 × 0.07)/3²
F = 37.8 × 10^(6) N
Answer:
a=2.304×10¹⁶m/s²
Explanation:
Given data
Distance d=2.5 nm=2,5×10⁻⁹m
Mass of proton m=1.6×10⁻²⁷kg
charge of proton q=1.6×10⁻¹⁹C
To find
acceleration a
Solution
Apply the Coulombs Law
Where k is coulombs constant (k=9×10⁹Nm²/C²)
q=q₁=q₂
r=d
So
Answer:
11.87 ms⁻¹
Explanation:
You can use the kinematic equation
v² = u² + 2as
Where v = final velocity
u = initial velocity
a = acceleration
s = displacement
v² = 9² + 2×1.5×20
So you get v = 11.87 ms⁻¹
<span>8,480 Joules. You must multiply by the angle in radians through which the target object rotates. 4.5 x 2 x 3.14 = 28.26666666 J for each turn
28.26666666 x 300 = 8480</span>