Answer:
125pi W/m
Explanation:
This is the formula for heat transfer by convection
Q=hA(T2-T1)
Q=Power
h=convection coefficient
A=area
T2= temperature of melting point
T1=temperature of air
Area of tow rods
A=2*pi*D=2*pi*0.01=pi/50
Q=10*(´pi/50)(650-25)
Q=125pi W/m
Answer:
Instrument Landing System
Explanation:
The ILS works by sending radio waves from the runway to the aircraft. Which is then intercepted and is used to guide the aircraft onto the runway.
Answer:
There are n types of coupons. Each newly obtained coupon is, independently, type i with probability p i , i = 1,...,n. Find the expected number and the variance of the number of distinct types obtained in a collection of k coupons
Explanation: The solution for the expectation has already been given in the comments. The calculation is slightly simplified by considering the number. Y= n-X of coupon types not collected with E[X] =N-E[Y] and Var(X) = Var(Y). Let Yi denote the indicator variable
Type i is not obtained, its expectations is the probability (1-pi)^k
Of not obtaining a coupon of type i, so the expected number of coupons is not obtained is
E[Y]= E{£iYi}= Ei(1-pi)^k
The variance is calculated analogously by expressing in terms of expectation,
Var(Y)=E{[EiYi]}^2-E[EiYi]^2
= Ei(1-pi)^k(1-(1-pi)^k + Ei=/j(1-pi-pj)^k-(1-pi)^k(1-pj)^k
Answer:
mg = 30.415 lbf
Explanation:
from figure body of size 3ft*2ft is tend to move down side
weight is divided into two component
vertical component = mgcos15 and
horizontal component = mg sin15
considering horizontal component equal to shear force
mgsin15 = \tau A
mgsin15 =\mu \frac{dv}{dh} A
mg =\frac{ \mu v A}{h*sin15}
=\frac{8.2*10^{-2}*0.2*3*2}{0.0125*sin15}
mg = 30.415 lbf
Answer:
Technician B is correct
Explanation:
To resize a connecting rod’s big end, the bolts are removed and the cap and rod mating surfaces will be ground to an approximate measurement. Thereafter, the rod cap will be bolted back onto the rod while the bore is machined back to it's proper size. However, the challenge with powdered metal rods is that due to the fact that the fractured surface is critical to their alignment, they cannot tolerate having those surfaces ground smooth. Hence, they cannot be resized, except bearing inserts of larger diameters are available for that particular engine.
Thus, technician B is correct.