Answer:
For number 4: A vector pointing to the right with a magnitude of 2.0
Explanation:
Very simple- just subtract 6-2
I am not sure how to do #2- sorry!
Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
Sodium, magnesium, and aluminum!
The amount of work done in emptying the tank by pumping the water over the top edge is 163.01* 10³ ft-lbs.
Given that, the tank is 8 feet across the top and 6 feet high
By the property of similar triangles, 4/6 = r/y
6r = 4y
r = 4/6*y = 2/3*y
Each disc is a circle with area, A = π(2/3*y)² = 4π/9*y²
The weight of each disc is m = ρw* A
m = 62.4* 4π/9*y² = 87.08*y²
The distance pumped is 6-y.
The work done in pumping the tank by pumping the water over the top edge is
W = 87.08 ∫(6-y)y² dy
W = 87.08 ∫(6y³ - y²) dy
W = 87.08 [6y⁴/4 - y³/3]
W = 87.08 [3y⁴/2- y³/3]
The limits are from 0 to 6.
W = 87.08 [3*6⁴/2 - 6³/3] = 87.08* [9*6³ - 2*36] = 87.08(1872) = 163013.76 ft-lbs
The amount of work done in emptying the tank by pumping the water over the top edge is 163013.76 ft-lbs.
To know more about work done:
brainly.com/question/16650139
#SPJ4
