I’m not sure with the question
Answer:
The acceleration of the woman is 0.44 m/s²
Explanation:
Given;
mass of the woman, m₁ = 90 kg
mass of the boy, m₂ = 60 kg
The force applied by the boy, f₂ = 40 N
The net horizontal force on the woman = 40 N
Apply Newton's second law of motion to determine the acceleration of the woman;
f = ma
a = f / m
a = 40 / 90
a = 0.44 m/s²
Therefore, the acceleration of the woman is 0.44 m/s²
Answer:
As the area of contact is less in the case of rolling than in the case of sliding, rolling friction is less than the sliding friction.
1. Frequency: 
The frequency of a light wave is given by:
where
is the speed of light
is the wavelength of the wave
In this problem, we have light with wavelength
Substituting into the equation, we find the frequency:
2. Period: 
The period of a wave is equal to the reciprocal of the frequency:
The frequency of this light wave is (found in the previous exercise), so the period is:
