A 90 k g shopper is standing on an escalator that makes a 25 ∘ incline with the horizontal. The escalator carries the shopper to the next floor, a height 30 m above ground level, at a constant speed. Calculate the work that the escalator does on the shopper.

W= 27 k J

Step-by-step explanation:

Given that

mass ,m = 90 kg

Height , h= 30 m

angle , θ = 25°

The work done by escalator is given as follows

W  = m g h

Work done by escalator does not depends on the angle ,it depends only on  height.

Now by by putting the values in the above equation we get

W= 90 x 10 x 30 J            ( take g = 10 m/s² )

W= 27000 J

W= 27 k J

Therefore work done by escalator will be 27 k J.

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.