120 km/h = 33.33 m/s 10.8 km/h/s = 3 m/s/s the motorist will cover a distance of 33.33 m x T(seconds) the police officer will cover a distance of 1/2 (3) T^2 they will be at the same point when 33.33T = 1.5 T^2 33.33 = 1.5 T 3T = 66.66 T = 22.22 seconds it will take the officer 22.22 seconds to catch the motorist. the officer will be moving V=AT = 3m/s x 22.22 seconds = 66.66 m/s almost 240 km/h (239.976 km/h)
The answer is is 730.5 days rounded, so A. 730
Answer:
Distance will be 49.34 m
Explanation:
We have given wavelength 
Diameter of the antenna d = 2.7 m
Range L = 7.8 km = 7800 m
We have to find the smallest distance hat two speedboats can be from each other and still be resolved as two separate objects D
We know that distance is given by 
So distance D will be 49.34 m
The solution for this problem:
The distance in mega parsec is equal to recession velocity / H, where h is equal to 50 mega parsec.The explanation for this is:1 parsec = 2.E+05 AU, nearly. 50 mega parsec = 1. E+13 AU, nearly. 1 mega means E+06 (million).
Answer: Rr = 900w
T = 1.66×10^-5s
P = 36.92W
Explanation: please find the attached file for the solution.