Best answer will be 70

13. the speed of light in a material is calculated by dividing the speed of light in a vacuum by the index of refraction in the material.

calculate the speed of light in water which has a refraction rate of 1.33. showing your work is not necessary.

14. the speed of light in a material is calculated by dividing the speed of light in a vacuum by the index of refraction in the material.

calculate the speed of light in glass which has a refraction rate of 1.50

showing your work is not necessary.

In a wind tunnel the speed changes as the cross sectional area of the tunnel changes. if the speed in a 6' x 6' square test section is 100 mph, what was the speed upstream of the test section where the tunnel measured 20' x 20'? use conservation of mass and assume incompressible flow. conservation of mass requires that as the flow moves through a path or a duct the product of the density, velocity and cross sectional area must remain constant; i.e., that ova-constant. a model is being tested in a wind tunnel at a speed of 100 mph if the flow in the test section is at sea level standard conditions, what is the pressure at the model's stagnation point? (a) the tunnel speed is being measured by a pitot-static tube connected to a u- tube manometer. what is the reading on that manometer in inches of water? (b) at one point on the model a pressure of 2058 psf is measured. what is the local airspeed at that point?

The speed of light can be used to measure the

Amass m = 74 kg slides on a frictionless track that has a drop, followed by a loop-the-loop with radius r = 18.4 m and finally a flat straight section at the same height as the center of the loop (18.4 m off the ground). since the mass would not make it around the loop if released from the height of the top of the loop (do you know why? ) it must be released above the top of the loop-the-loop height. (assume the mass never leaves the smooth track at any point on its path.) 1. what is the minimum speed the block must have at the top of the loop to make it around the loop-the-loop without leaving the track? 2. what height above the ground must the mass begin to make it around the loop-the-loop? 3. if the mass has just enough speed to make it around the loop without leaving the track, what will its speed be at the bottom of the loop? 4. if the mass has just enough speed to make it around the loop without leaving the track, what is its speed at the final flat level (18.4 m off the ground)? 5. now a spring with spring constant k = 15600 n/m is used on the final flat surface to stop the mass. how far does the spring compress?