17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106
1 answer:
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Answer:
change in height is 1.664 mm
Explanation:
Given data
drops = 3.00 mm
diameter = 5.00 cm = 0.05 mm
decrease = 350 cm^3
temperature = 95°C to 44.0°C
to find out
the decrease in millimeters in level
solution
we will calculate here change in volume so we can find how much level is decrease
change in volume = β v change in temp ...............1
here change in volume = area× height
so = /4 × d² h
so we can say change in volume = /4 × d² × change in height .......2
so from equation 1 and 2 we calculate change in height
( β(w) -β(g) )× v× change in temp = /4 × d² × change in height
change in height = 4 × ( β(w) -β(g) ) v× change in temp / /4 × d²
put all value here
change in height = 4 × ( 210 - 27 )(350 ) × (95-44) /
/4 × 0.05²
change in height is 1.664 mm
Explanation:
The 11Ω, 22Ω, and 33Ω resistors are in parallel. That combination is in series with the 4Ω and 10Ω resistors.
The net resistance is:
R = 4Ω + 10Ω + 1/(1/11Ω + 1/22Ω + 1/33Ω)
R = 20Ω
Using Ohm's law, we can find the current going through the 4Ω and 10Ω resistors:
V = IR
120 V = I (20Ω)
I = 6 A
So the voltage drops are:
V = (4Ω) (6A) = 24 V
V = (10Ω) (6A) = 60 V
That means the voltage drop across the 11Ω, 22Ω, and 33Ω resistors is:
V = 120 V − 24 V − 60 V
V = 36 V
So the currents are:
I = 36 V / 11 Ω = 3.27 A
I = 36 V / 22 Ω = 1.64 A
I = 36 V / 33 Ω = 1.09 A
If we wanted to, we could also show this using Kirchhoff's laws.
