Answer:
0.0659 A
Explanation:
Given that :
( saturation current )
at 25°c = 300 k ( room temperature )
n = 2 for silicon diode
Determine the saturation current at 100 degrees = 373 k
Diode equation at room temperature = I = Io 
next we have to determine the value of V at 373 k
q / kT = (1.6 * 10^-19) / (1.38 * 10^-23 * 373) = 31.08 V^-1
Given that I is constant
Io =
= 0.0659 A
Answer:
s= 20.4 m
Explanation:
First lets write down equations for each ball:
s=so+vo*t+1/2a_c*t^2
for ball A:
s_a=30+5*t+1/2*9.81*t^2
for ball B:
s_b=20*t-1/2*9.81*t^2
to find time deeded to pass we just put that
s_a = s_b
30+5*t-4.91*t^2=20*t-4.9*t^2
t=2 s
now we just have to put that time in any of those equations an get distance from the ground:
s = 30 + 5*2 -1/2*9.81 *2^2
s= 20.4 m
Answer:
Recognize that there is a moral dilemma.
Determine the actor. ...
Gather the relevant facts. ...
Test for right versus wrong issues. ...
Test for right versus right paradigms. ...
Apply the resolution principles. ...
Investigate the trilemma options. ...
Make the decision.
Answer:
5984.67N
Explanation:
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?
from continuity equation
v1A1=v2A2
equation of continuity
v1=4ft /s=1.21m/s
d1=14 inch=.35m
d2=14-2=0.304m
A1=pi*d^2/4
0.096m^2
a2=0.0706m^2
from continuity once again
1.21*0.096=v2(0.07)
v2=1.65
force on the pipe
(p1A1- p2A2) + m(v2 – v1)
from bernoulli
p1 + ρv1^2/2 = p2 + ρv2^2/2
difference in pressure or pressure drop
p1-p2=2psi
13.789N/m^2=rho(1.65^2-1.21^2)/2
rho=21.91kg/m^3
since the pipe is cylindrical
pressure is egh
13.789=21.91*9.81*h
length of the pipe is
0.064m
AH=volume of the pipe(area *h)
the mass =rho*A*H
0.064*0.07*21.91
m=0.098kg
(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)
force =5984.67N