Both A and B technicians are correct because both might be used to test fuses, according to technician B.
<h3>What is continuity?</h3>
The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.
From the question:
We can conclude:
The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.
Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.
Technician A says both an ohmmeter and a self-powered test light may be used to test for continuity. Technician B says both may be used to test fuses. Who is correct?
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Answer:
M =2.33 kg
Explanation:
given data:
mass of piston - 2kg
diameter of piston is 10 cm
height of water 30 cm
atmospheric pressure 101 kPa
water temperature = 50°C
Density of water at 50 degree celcius is 988kg/m^3
volume of cylinder is 


mass of available in the given container is



M =2.33 kg
Answer:
The glass transition occur when the given temperature gets drop below the temperature of the glass transition for the polymers. During the creases of the long range of the given motion, the polymer start changing from the state of hard into the form of brittle.
The range of the glass transition temperature are between the 0 to -150°C. The basic use of the polymer is basically depend upon the main properties of the polymer.
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
Explanation:!!
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