If a dune buggy drives from an elevation of 10 meters below sea level to an elevation of 5 meters above sea level, what is the vertical distance that the dune buggy traveled?

5

Step-by-step explanation:

To find the answer you would subtract the 10 meters minus the 5 meters and then you get your result which is 5.

(sin 42)/31. = sinx)/37

sinx = 37 (sin42)/31

= 37× 0.66/31 = 0.787

x= 51.9

other angle opposite to y = 180-51.9-42

= 180- 93.9 =86.1

so

sin42)/31 = sin 86.1)/y

0.021 = 0.997 /y

y= 0.997/0.021 =47.47

a. 12.5 kg

step-by-step explanation:

f = ma

so

m = f/a = (2500 n)/(200 m/s²) = 12.5 kg

1 n = 1 kg·m/s²