Twenty is the atomic number of potassium.
Answer:
a = ω^2 A formula for max acceleration (ignoring sign)
V = ω A formula for max velocity
V^2 = ω^2 A^2 = a A from first equation
E = 1/2 M V^2 = 1/2 * 2.98 * 3.55 * .0805 = .426 J
(kg * m/sec^2 * m = kg m^2 / sec^2 = Joule
Hi! the atom in this particular problem has emitted an alpha particle in a nuclear reaction.
Glad I could help, and happy learning!
Answer:
Magnitude of static friction force is 70 sin40° = 44.99 N.
No, it is not necessary that it is maximum static friction.
Normal force is equal to 70 cos40° = 53.62 N.
Explanation:
We apply newton law of moton equation along the plane and perpendicular to plane;
Along the plane,
70 sin 40° =
---------------(1)
70 cos 40° = N --------------(2)
= μN -----------------(3)
So, it depends on the value of μ that the friction is maximum or not .