Answer:I would prefer ginger with lemon and honey tea...Stay hydrated,Rest,
Explanation:
Answer:
A
Explanation:
If you do B, the wagon slows down due to friction
If you do C, it would slow down the wagon by half
If you do D, it would slow it down, due to there being more weight but same force.
Answer:
Average :
UCL = 4.15
LCL = 2.65
Range :
UCL = 2.75
LCL = 0
Explanation:
Given :
Sample size, n = 5
Average, X = 3.4
Range, R = 1.3
A2 for n = 5 ; equals 0.577 ( X chart table)
For the average :
Upper Control Limit (UCL) :
X + A2*R
3.4 + 0.577(1.3) = 4.1501
Lower Control Limit (LCL) :
X - A2*R
3.4 - 0.577(1.3) = 2.6499
FOR the range :
Upper Control Limit (UCL) :
UCL = D4*R
D4 for n = 5 ; equals = 2.114
UCL = 2.114*1.3 = 2.7482
Lower Control Limit (LCL) :
LCL = D3*R
D3 for n = 5 ; equals = 0
LCL = 0 * 1.3 = 0
The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)
<span>To answer this problem, we use balancing of forces: x and y components to determine the tension of the rope.
First, the vertical component of tension (Tsin theta) is equal to the weight of the object.
T * sin θ = mg =</span> 1.55 * 9.81 <span>
T * sin θ = 15.2055
Second, the horizontal component of tension (t cos theta) is equal to the force of the wind.
T * cos θ = 13.3
Tan θ = sin </span>θ / cos θ = 15.2055/13.3 = 1.143
we can find θ that is equal to 48.82.
T then is equal to 20.20 N
