Answer:
The distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Explanation:
Given;
speed of the faster car, v₁ = 60 mi/h
speed of the slower car, v₂ = 55 mi/h
Let the distance traveled by the faster car when it is 15 mins ahead of the slower car = x miles
Note: divide 15 mins by 60 to convert to hours for consistency in the units.
Therefore, the distance traveled by the faster car when it is 15 mins ahead of the slower car is 165 miles.
Answer:
An investigation is made to determine the performance of simple thin airfoils in the slightly supersonic flow region with the aid of the nonlinear transonic theory first developed by von Kármán[1]. Expressions for the pressure coefficient across an oblique shock and a Prandtl-Meyer expansion are developed in terms of a transonic similarity parameter. Aerodynamic coefficients are calculated in similarity form for the flat plate and asymmetric wedge airfoils, and curves are plotted. Sample curves for a flat plate and a specific asymmetric wedge are plotted on the usual coordinate grid of Cl, Cd,andCmc/4versus angle of attack and Cl versus Mach Number to illustrate the apparent features of nonlinear flow.
Explanation:
Explanation:
0.566kg *(1mol/0.197 kg)= 2.87 mol gold
note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong
I think the logical question here is to either find the distance or the displacement. They differ in such a way that distance is a scalar quantity that does not focus on the direction. Displacement is a vector quantity that covers the distance from the starting point to end point. Because it travels only in one direction (to the east), in this condition, distance is equal to displacement.
Distance = Displacement = 3,000 m + 1,500 m = 4,500 m
D lung cancer is not infectious
