Answer:
3 sigma lower control limit = 0.0429
Explanation:
Given.
n = 100
days = 100
Number of defective bulbs = 600 defective bulbs
Let p = Process Average
p = 600/(100*40)
P = 600/4000
P = 0.15
q = 1 - p
q = 1 - 0.15
q = 0.85
3 sigma lower limit = p - 3*√(pq/n)
Using the above formula
Substitute in the values
3 sigma lower control limit = 0.15 - 3 * √(0.15 * 0.85/100)
3 sigma lower control limit= 0.15 - 3√0.001275
3 sigma lower control limit = 0.15 - 3* 0.035707142142714
3 sigma lower control limit = 0.15 - 0.107121426428142
3 sigma lower control limit = 0.04287857357185
3 sigma lower control limit = 0.0429 ---- approximated
Answer: Magnitude of the force exerted on the egg by the ground is 9.2N
Explanation:
Given the following :
Mass of egg (m) = 150g = 0.15kg
Height(h) from which egg is dropped = 3m
velocity of egg before hitting the ground (u) = 4.4m/s
Final velocity of egg (V) = 0
Time taken (t) = 0.072s
Magnitude of the force exerted on the egg by the ground can be found by applying Newton's 2nd law:
Momentum = mass × velocity
From Newton's second law:
Force = mass × change in Velocity with time ;
That is
F = m * ΔV / t
Inputting our values
F = 0.15 * (4.4 - 0) / 0.072
F = 0.15 × (4.4 / 0.072)
F = 0.15 × 61.11
F = 9.16N
F = 9.2N
Answer:
SECOND LAW OF NEWTON
Explanation:
When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression
I = F t = Δp
As we can see this expression is a form of Newton's second law
F = m a
a = dv / dt
F = m dv / dt
F dt = m dv
p = mv
F dt = dp
Therefore the station moves through the SECOND LAW OF NEWTON
Answer:
60*12.0= 720 = v/60 * 12.0 squared which is 1,728
Explanation:
Horizontal velocity component: Vx = V * cos(α)
