Answer:
0.466 (3 sig. fig.)
Explanation:
Frictional force acting on the box = 5.00×10^2xsin25
Normal force acting on the box = 5.00×10^2xcos25
coefficient of friction = 0.466 (3 sig. fig.)
7500m because 1,500 in 1 sec so 1500x5 7500
P=I*E
Power (P)
Voltage (E)
Amps (I)
If the box is a distance 1.81 m from the rear of the truck when the truck starts,<span> ... Force of Friction = mu_s * Normal Force( </span>M<span> * G) ... The </span>box starts<span> moving! ... Now that the </span>box<span> is moving, the bed of the </span>truck<span> pulls at it with 17.4 ... out how </span>long<span> it will take the </span>box<span> to reach the back of the </span>truck<span>. ... T^2 = 2 * </span>1.81<span> / .64</span>