The equation to be used is for the rectilinear motion at constant acceleration:
x = v₀t + 0.5at²
a = (v-v₀)/t
where
x is distance
v and v₀ is the final and initial velocity
t is time
a is acceleration
Because the acceleration is decelerating, that would be reported as -7.5 m/s². Substituting,
-7.5 = (0 - v₀)/t
v₀ = 7.5 t --> eqn 1
x = v₀t + 0.5at²
60 = (7.5t)(t) + 0.5(-7.5)(t²)
Solving for t,
t = 4s
Thus,
v₀ = 7.5 m/s² * 4s
v₀ = 30 m/s
Answer:
Kinetic friction is lesser than limiting friction. Two surfaces are rubbed together, first with a smaller force and then with a greater force.
Answer:
The earth's pull on the moon
Explanation:
Earth exerts a gravitational pull on the moon 80 times stronger than the moon's pull on the Earth.
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.
