Answer: 2.7 m/s
Explanation:
Given the following :
Period (T) = 8.2 seconds
Radius = 3.5 m
The tangential speed is given as:
V = Radius × ω
ω = angular speed = (2 × pi) / T
ω = (2 × 22/7) / 8.2
ω = 6.2857142 / 8.2
ω = 0.7665505
Therefore, tangential speed (V) equals;
r × ω
3.5 × 0.7665505 = 2.6829268 m/s
2.7 m/s
Answer:
The mass of the other worker is 45 kg
Explanation:
The given parameters are;
The gravitational potential energy of one construction worker = The gravitational potential energy of the other construction worker
The mass of the lighter construction worker, m₁ = 90 kg
The height level of the lighter construction worker's location = h₁
The height level of the other construction worker's location = h₂ = 2·h₁
The gravitational potential energy, P.E., is given as follows;
P.E. = m·g·h
Where;
m = The mass of the object at height
g = The acceleration due to gravity
h = The height at which is located
Let P.E.₁ represent the gravitational potential energy of one construction worker and let P.E.₂ represent the gravitational potential energy of the other construction worker
We have;
P.E.₁ = P.E.₂
Therefore;
m₁·g·h₁ = m₂·g·h₂
h₂ = 2·h₁
We have;
m₁·g·h₁ = m₂·g·2·h₁
m₁ = 2·m₂
90 kg = 2 × m₂
m₂ = (90 kg)/2 = 45 kg
The mass of the other construction worker is 45 kg.
If the two waves have the SAME FREQUENCY and are exactly
out of phase (180° apart), then the resultant wave will have the
same frequency and an amplitude of 1 unit.
If the two waves do not have the SAME FREQUENCY, then their
relative phase is meaningless.
