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
Answer:
The wheelbarrow's wheel and axle help the wheelbarrow to move without friction thus making it easier to push or pull. That's why it will be easier to lift a load in wheel barrow of the load is transferred towards the wheel.
Answer: 0.258 N
Explanation:
As the density of the object is much less than the density of water, it’s clear that the buoyant force, is greater than the weight of the object, which means that in normal conditions, it would float in water.
So, in order to get the ball submerged in water, we need to add a downward force, that add to the weight, in order to compensate the buoyant force, as follows:
F = Fb – Fg
Fb= δH20* 4/3*π*(d/2)³ * g
Fg = δb* 4/3*π*(d/2)³ *g
F= (δH20- δb) * 4/3*π*(d/2)³*g
Replacing by the values of the densities, and the ball diameter, we finally get:
F= 0.258 N
Answer:
2N
Explanation:
subtract rthe two forces to see which is greater
4-2=2
Answer:
v = 12.12 m/s
Explanation:
Given that,
The mass of the cart, m = 75 kg
The roller coaster begins 15 m above the ground.
We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m
So, the velocity of the cart is 12.12 m/s.
