A610-n hiker carrying an 11.0-kg backpack hiked up a trail for 23 minutes. at the end of that time, he is 150 m higher than when he started. (a) how much the the backpack weigh? (b) how much work is accomplished to carry the backpack that distance? (c) how much total work is accomplished? (use the total weight: hiker + backpack.) (d) what is the hiker’s power in watts? (don’t forget to convert 23 minutes to seconds.)

a)110 N (assuming g = 10 m/s2 )

b) 16500 J

c) ≅ 11.96 W

Explanation:

a) 150 m is not enough to a sensible change of gravitational acceleration at that level from Earth's surface. Therefore, backback weight is same as when it is in ground level.

W = mg, W = weight of object ; m = mass of the object ; g =  gravitational acceleration felt by object.

By substituting,

W = 11×10 = 110 N (assuming g = 10 m/s2 )

b)  Work done = Energy change

    Work done by lifting the bag = Potential energy change

                                                     = mgh ; h = Vertical height lifted

                                                     = 11×150×10 (assuming g = 10 m/s2 )

                                                     = 16500 J

c) Power = Work done/ time taken

               = 16500/(23×60)                (minutes converted to seconds by multiplying by 60)

               ≅ 11.96 W

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2