A child fills a bucket with sand so that the bucket and sand together weigh 10 lbs, lifts it 2 feet up and then walks along the beach, holding the bucket at a constant height of 2 ft above the ground. How much work is done on the bucket after the child has walked 100 ft?

Total weight, w = 10 lbs.

Height of the bucket, h = 2 feet.

Distance walked, d = 100 ft.

Now, work done in moving the bucket at a height of 2 feet.

W = mgh

W = 2× 32.17×100

W = 6434  lbs ft²/s²

Work done in moving bucket in horizontal direction is zero because it is perpendicular to the force.

Therefore, work done is 6434  lbs ft²/s² .

Hence, this is the required solution.

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation:

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.