D. Using a fixed-pulley system
Answer:
L = 2.8 cm
Explanation:
Period T = 4 / 12 = 1/3 s
T = 2π√(L/g)
L = (T/2π)²g
L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm
RADIATION BELTS....... I think but it should be radiation belt
<span>We can use Coulomb's law to find the force F acting on the proton that is released.
F = k x Q1 x Q2 / r^2
k = 9 x 10^9
Q1 is the charge on one proton which is 1.6 x 10^{-19} C
Q2 is the same charge on the other proton
r is the distance between the protons
F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2
F = 2.304 x 10^{-22} N
We can use the force to find the acceleration.
F = ma
a = F / m
a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg)
a = 1.38 x 10^5 m/s^2
The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
Answer:
Im gonna say it is answer A:) Hope this helps!
Explanation:
