If a rock is thrown off the side of a cliff 80m high with an initial speed of 30 m/s in the horizontal direction, what is the rock's speed at the bottom of the cliff (just before impact)?
Since the initial speed is entirely horizontal, finding the vertical component of the speed at the bottom has the initial condition of starting at 0 m/s in the vertical direction. Therefore using vf^2=vo^2 + 2gy, where vo is zero, g is 10 m/s^2, and y is 80m, vf is calculated then to be 40m/s in the vertical direction. As for the horizontal component of the rock's velocity, there is no horizontal acceleration of the rock (ignoring air resistance), making the final horizontal speed still 30m/s. To find the resultant speed, the horizontal and vertical speeds must be added vectorally, which can be done with the pythagorean theorem since they form a right triangle. Therefore v^2=30^2 + 40^2, leaving the final speed at 50 m/s.
A Point charge of charge +Q resides in the center of a hollow regular tetrahedron of side length a. Find the outward flux through one of the faces of the tetrahedron.
Using Gauss's Law, the total flux from an enclosed charge is equal to q/ε. Since the tetrahedron symmetrically encloses the point charge, we can say that 1/4th of the total flux goes through each face of the shape, since there are 4 faces in a tetrahedron. From this, the flux through one face must be (+Q/ε)*(1/4)= +Q/(4ε) where ε is the permittivity of free space.
Using the Divergence Theorem, find the flux of the velocity field F=<x,y,z> through the surface f(x,y)=4-x^2-y^2 bounded by the x-y plane.
The divergence Theorem states that the flux surface integral of a vector valued function over a closed surface is equivalent to the Volume integral of the divergence of the function over the volume enclosed by the given surface. ∯〖F(x,y,z)dS 〗=∰∇⋅F dV The divergence of this function is the sum of the partial derivatives with respect to each variable, so 1+1+1=3. Evaluating the triple integral where z ranges from 0 to 4-x^2-y^2, and x and y are converted to polar coordinates, r ranges from 0 to 2 and θ ranges from 0 to 2π. Evaluating this triple integral gives ∬3(4-r^2 )rdrdθ. Evaluating the next integral gives 12∫_0^2π▒dθ=24 π.