The velocity of the ball when it reaches the ground is equal to B. 68.6 m/s. This value was obtained from the formula Vf = Vi + at. Vf is the final velocity. Vi is the initial velocity. The acceleration is "a", while the time of travel is "t". The solution is:
<span>Vf = Vi + at
</span>Vf = 0 + (-9.8 m/s^2) (7 s)
Vf = -68.6 m/s
The negative sign denotes the direction of the ball.
They can either cancel each other or add up to a resultant force with a certain direction and modulus.
Newton's second law states that F=m*a, where F is the resultant force, ie ΣF.
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.
Refer to the diagram shown below.
Let I = the moment of inertia of the wheel.
α = 0.81 rad/s², the angular acceleration
r = 0.33 m, the radius of the weel
F = 260 N, the applied tangential force
The applied torque is
T = F*r
= (260 N)*(0.33 m)
= 85.8 N-m
By definition,
T = I*α
Therefore,
I = T/α
= (85.8 N-m)/(0.81 rad/s²)
= 105.93 kg-m²
Answer: 105.93 kg-m²