In a shipping yard, a crane operator attaches a cable to a 1,090 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 32 m. Determine the amount of work done (in J) by each of the following.

a. the tension in the cable
b. the force of gravity

It may possibly be c i think |-/

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.

Answer sample (c); // would be best because the percentage of the energy in an incident wave that remains in a reflected wave from this material is the smallest;