<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>
Answer:
1.90×10²⁰ Electrons
Explanation:
From the question,
Q = It.................... Equation 1
Where Q = charge flowing through the wire, I = current, t = time
Given: I = 4.35 A, t = 7.00 s
Substitute these values into equation 1
Q = 4.35(7.00)
Q = 30.45 C.
But,
1 electron contains 1.6×10⁻¹⁹ C
therefore,
30.45 C = 30.45/1.6×10⁻¹⁹ electrons
= 1.90×10²⁰ Electrons
It is a mechanical wave and cannot travel through a vacuum.
