How can shoe and tire marks be preserved in mud?

<h2>RED!</h2><h3></h3><h3>On the visible spectrum, red has the lowest frequency.</h3><h3>(I'm an amateur astronomer, so I would know.)</h3>

4 0

2 years ago

An airplane has a speed of 135 mi/h in still air. It is flying straight north so that it is always directly above a north-south

vlabodo [156]

Answer:

The answer is below

Explanation:

Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.

The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph

vₐ = vₙ + vₐₙ

vₐ = vₐₙ

Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.

The direction of the wind θ is:

sin(θ / 2) = vₙ / 2vₐ

sin(θ / 2) = 70/ (2*135)

sin(θ / 2) = 0.2593

θ / 2 = sin⁻¹(0.2593) = 15

θ = 30⁰

2α = 180° - 30°

2α = 150°

α = 75°

a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.

8 0

3 years ago

Ml(d^2θ/dt^2) =-mgθ

Nata [24]

The equation of motion of a pendulum is:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,$

where $\ell$ it its length and $g$ is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles ( $\theta \ll 1$ ), we can use:

$\sin\theta \simeq \theta.$

Additionally, let us define:

$\omega^2\equiv\dfrac{g}{\ell}.$

We can now write:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.$

The solution to this differential equation is:

$\theta(t) = A\sin(\omega t + \phi),$

where $A$ and $\phi$ are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.$

This justifies that the period depends only on the pendulum's length.

4 0

3 years ago