Kinetic energy = (1/2) (mass) (speed)²
= (1/2) (1.4 kg) (22.5 m/s)²
= (0.7 kg) (506.25 m²/s² )
= 354.375 kg-m²/s² = 354.375 joules .
This is just the kinetic energy associated with a 1.4-kg glob of
mass sailing through space at 22.5 m/s. In the case of a frisbee,
it's also spinning, and there's some additional kinetic energy stored
in the spin.
Answer:
T = 3.23 s
Explanation:
In the simple harmonic movement of a spring with a mass the angular velocity is given by
w = √ K / m
With the initial data let's look for the ratio k / m
The angular velocity is related to the frequency and period
w = 2π f = 2π / T
2π / T = √ k / m
k₀ / m₀ = (2π / T)²
k₀ / m₀ = (2π / 3.0)²
k₀ / m₀ = 4.3865
The period on the new planet is
2π / T = √ k / m
T = 2π √ m / k
In this case the amounts are
m = 6 m₀
k = 10 k₀
We replace
T = 2π√6m₀ / 10k₀
T = 2π √6/10 √m₀ / k₀
T = 2π √ 0.6 √1 / 4.3865
T = 3.23 s
Answer:
30N*s
Explanation:
Given the following data;
Force = 10N
Time = 3 seconds
To find the impulse;
Impulse = force * time
Substituting into the equation, we have;
Impulse = 10 * 3
Impulse = 30Ns
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.