Explanation:
Given:
m = 1.673 × 10^-27 kg
Q = q = 1.602 × 10^-19 C
r = 0.75 nm
= 0.75 × 10^-9 m
A.
Energy, U = (kQq)/r
Ut = 1/2 mv^2 + 1/2 mv^2
1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9
v = 1.356 × 10^4 m/s
B.
F = (kQq)/r^2
F = m × a
1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2
a = 2.45 × 10^17 m/s^2.
Answer:
The solution is given in the picture attached below
Explanation:
200 joules of work energy are involved. That's all we need to know to answer the question. Once we know that 200 joules of work energy are involved, we don't care what was lifted, or how far, or how long it took, or how many people worked on it, or how much they were paid, or what was the distribution of their gender identities, or the ethnic diversity among the team. or what day each of them celebrates as their sabbath. Any other information besides the 200 joules is only there to distract us, and see whether we're paying attention.
Power = (work or energy) / (time to do the work or move the energy)
Power = (200 joules) / (5 seconds)
<em>Power = 40 watts</em>
Yes for an object moving on a horizontal plane, R = mg (where mg = weight). therefore, for an object moving on a horizontal plane: F = μmg
