let Coefficients of Friction of Rubber on asphalt (dry) =0.7
F= Coefficients of Friction * normal force = 0.7 * 60 =42 N
so the net force of the rubber is zero, meaning it will travel at a constant speed.
When the rubber is travel at 2m/s, 42N is required to keep moving at constant speed
Answer:
Part a)

Part b)

Explanation:
Part a)
Level of sound = 75 dB
now we know that

here we know that

now we have


Part b)
Intensity of sound wave is given as

here we know that

so we have


now we know



now we have


Answer:
The rise from A to B is 0.887
Solution:
As per the question:
The following reading of an inverted staff is given as:
A = 2.915
B = -2.028
Here, for inverted staff, the greater reading shows greater elevation and lesser reading shows lower elevation.
Thus
The rise from A to B is given as:
A - B = 2.915 - 2.028 = 0.887
Am sorry what can you be more specific
Answer:
V₁ = √ (gy / 3)
Explanation:
For this exercise we will use the concepts of mechanical energy, for which we define energy n the initial point and the point of average height and / 2
Starting point
Em₀ = U₁ + U₂
Em₀ = m₁ g y₁ + m₂ g y₂
Let's place the reference system at the point where the mass m1 is
y₁ = 0
y₂ = y
Em₀ = m₂ g y = 2 m₁ g y
End point, at height yf = y / 2
= K₁ + U₁ + K₂ + U₂
= ½ m₁ v₁² + ½ m₂ v₂² + m₁ g
+ m₂ g 
Since the masses are joined by a rope, they must have the same speed
= ½ (m₁ + m₂) v₁² + (m₁ + m₂) g 
= ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
How energy is conserved
Em₀ = 
2 m₁ g y = ½ (m₁ + 2m₁) v₁² + (m₁ + 2m₁) g 
2 m₁ g y = ½ (3m₁) v₁² + (3m₁) g y / 2
3/2 v₁² = 2 g y -3/2 g y
3/2 v₁² = ½ g y
V₁ = √ (gy / 3)