Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity 
= 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e ₂ = ₁/2
Hence,
₂/₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
When the child is moving, he/she has kinetic energy. For just a brief second before they move the other way, the child is not moving, but they have gravitational potential energy.
The child may need a push from time to time because friction with the air causes loss of energy.
To solve this problem we will apply the concepts related to the balance of forces. We will decompose the forces in the vertical and horizontal sense, and at the same time, we will perform summation of torques to eliminate some variables and obtain a system of equations that allow us to obtain the angle.
The forces in the vertical direction would be,
The forces in the horizontal direction would be,
The sum of Torques at equilibrium,
The maximum friction force would be equivalent to the coefficient of friction by the person, but at the same time to the expression previously found, therefore
Replacing,
Therefore the minimum angle that the person can reach is 46.9°
Answer:
Assuming there are 28 days in each month,
750W = 0.75kW
Cost of electric bill = 0.75 × 8 × 28 × $0.23
= $38.64
