Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
Answer:

Explanation:
c = Speed of wave
= Density of medium
A = Area
= Frequency

Intensity of sound is given by

So,

We get

The ratio is 
We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.
T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F
To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as
F = ma
Where,
m= Mass
a = Acceleration
At the same time the frictional force can be defined as,

Where,
Frictional coefficient
N = Normal force (mass*gravity)
Our values are given as,

By condition of Balance the friction force must be equal to the total net force, that is to say



Re-arrange to find acceleration,



Therefore the acceleration the horse can give is 