Please answer as many questions as possible, I want to double check my answers.
ID: 2271653 • Letter: P
Question
Please answer as many questions as possible, I want to double check my answers. Thanks
A truck travels at an average velocity of 250 fpm from a loader to a conveyor where it will discharge its payload. If travel time is 70 sec. what is the distance between the loading and dumping points? Answer: A mine locomotive accelerates from a velocity of 3 mph to 6 mph in 45 sec. What is the rate of acceleration in ft/sec2? Answer: If a truck accelerates at a uniform rate of 0.5 mph sec from rest, how fast will it be going in 25 sec? Answer: What is the average velocity for a truck, in Problem 3. during the time interval between its initial velocity at 25 sec. 12.5 mph, and its final velocity at 1 min. 30 mph? Answer: How far does the truck travel in Problem 4? Answer: A 10-ton vehicle is traveling at 6 mph when the driver sees a falling rock ahead of him. He applies the brakes but still hits the rock moving it 4 inches. What is the force of impact in lb force? Answer: Hint The force of impact (work done) is equal to die change of kinetic energy. A 20-ton vehicle is traveling at 6 mph. Determine the retarding force, in lb force of the brakes required to stop it in 100 ft on level. Answer: Hint Use the Newton's law A rock weighs 160 lb ft3. What is its density? What is its specific gravity (S G)? What is its tonnage factor (TF)? What would be its density. SG and TF in metric system? Answer: Bituminous coal has a swell factor of 0.74. What is the volume of blasted coal, ready for loading if its bank volume was 1200 ft3? Answer: Calculate both bank and loose densities (in lb ft3) of the coal in problem 9 if the loose tonnage factor (TF) is 42.9ft3/ton. Calculate both densities in metric system. Answer: What is the %swell for coal in Problem 10? Answer: Calculate bank volume of 37.8 tons of coal if its loose density is 46.6 lb ft3 and %swell as in Problem 11. Answer:Explanation / Answer
1) 1 ft = 0.3048 m
d = v*t = 250*0.3048/60 * 70 = 88.9 m
2) v1 = 3 mph = 3*1.6*1000/3660 = 1.333 m/s
v2 = 6 mph = 2.666 m/s
a = (v2-v1)/t = (2.666-1.333)/45 = 0.02963 m/s^2 = 0.02963/0.3048 = 0.0972 ft/s^2
3) a = 0.5*1.6*1000m/(3600s^2) = 0.178 m/s^2
v = a*t = 0.178*25 = 4.45 m/s
4) s = 0.5*a*t^2 = 0.5*0.178*25^2 = 55.56 m
Vav = s/t = 55.56/25 = 2.22 m/s