1. Let us filled with grace.
2. Let us be filled with grace.
(Which one is correct? Do we have to add 'the' before 'grace?)
3 answers
2. is correct and you do not need "the" before "grace."
Hey "drwls" can you explain to me how you solved a previous physics question that you answered? It's for the following question below, and it's already posted on here. Unfortunately, it didn't allow me to post the link to it, so you may have to search for it. If you do find it, I have already posted what I'm exactly unsure of on that specific page. Thanks.
Here is the question:
A banked circular highway curve is designed for traffic moving at 55 km/h. The radius of the curve is 220 m. Traffic is moving along the highway at 38 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)
Here is the question:
A banked circular highway curve is designed for traffic moving at 55 km/h. The radius of the curve is 220 m. Traffic is moving along the highway at 38 km/h on a rainy day. What is the minimum coefficient of friction between tires and road that will allow cars to take the turn without sliding off the road? (Assume the cars do not have negative lift.)
I do not remember answering that question.
55 km/h = 15.28 m/s Call that V
If the car is banked for turns at that velocity, the normal force component in the vertical direction is Mg and the component in the hotizontal direction is M g tan A, which equals MV^2/R.
Thus the ideal bank angle for that velocity is is arctan V^2/(Rg) = 6.18 degrees.
If the car is travelling at a slower V' = 38 km/h = 10.56 m/s, static friction must provide enough force away from the center of the turn to keep the car from sliding toward the center.
It appears that two equations must be solved in two unknowns to get the required static friction coefficient. I do not have time to provide the solution for you
55 km/h = 15.28 m/s Call that V
If the car is banked for turns at that velocity, the normal force component in the vertical direction is Mg and the component in the hotizontal direction is M g tan A, which equals MV^2/R.
Thus the ideal bank angle for that velocity is is arctan V^2/(Rg) = 6.18 degrees.
If the car is travelling at a slower V' = 38 km/h = 10.56 m/s, static friction must provide enough force away from the center of the turn to keep the car from sliding toward the center.
It appears that two equations must be solved in two unknowns to get the required static friction coefficient. I do not have time to provide the solution for you
33 answers