Hence the expression of ω in terms of m and k is

Given the expressions;

Equating both expressions we will have;

Divide both equations by 2π

Square both sides

Take the square root of both sides

Hence the expression of ω in terms of m and k is

Speed, being a scalar quantity, is the rate at which an object covers distance. The average speed is the distance (a scalar quantity) per time ratio. ... Velocity is the rate at which the position changes. The average velocity is the displacement or position change (a vector quantity) per time ratio.
Answer:
Force, F = 187.42 N
Explanation:
It is given that,
Mass of boy, m = 30 kg
Acceleration due to gravity, 
Radius of curvature of the roller coaster, r = 15 m
Speed of the car, v = 7.3 m/s
The force acting on the boy are force of gravity and the centripetal force. The net force acting on him is as follows :



F = 187.42 N
So, he press against the seat with a force is 187.42 N. Hence, this is the required solution.
Answer:
15 m/s or 1500 cm/s
Explanation:
Given that
Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s
Distance moved during the hook, d(h) = 5 cm = 0.05 m
Distance moved by the fist, d(f) = 100 cm = 1 m
Average speed of the fist during the hook, v(f) = ? cm/s = m/s
This can be solved by a very simple relation.
d(f) / d(h) = v(f) / v(h)
v(f) = [d(f) * v(h)] / d(h)
v(f) = (1 * 0.75) / 0.05
v(f) = 0.75 / 0.05
v(f) = 15 m/s
Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s
Answer:
B. 6HgO → 6Hg + 3O
Explanation:
A decomposition reaction is a reaction in which a single reactant is broken down into 2 or more products.