emf generated by the coil is 1.57 V
Explanation:
Given details-
Number of turns of wire- 1000 turns
The diameter of the wire coil- 1 cm
Magnetic field (Initial)= 0.10 T
Magnetic Field (Final)=0.30 T
Time=10 ms
The orientation of the axis of the coil= parallel to the field.
We know that EMF of the coil is mathematically represented as –
E=N(ΔФ/Δt)
Where E= emf generated
ΔФ= change inmagnetic flux
Δt= change in time
N= no of turns*area of the coil
Substituting the values of the above variables
=1000*3.14*0.5*10-4
=.0785
E=0.0785(.2/10*10-3)
=1.57 V
Thus, the emf generated is 1.57 V
Answer:
hello your question is incomplete attached below is the complete question
A) overall mean = 5.535, standard deviation ≈ 0.3239
B ) upper limit = 5.85, lower limit = 5.0
C) Not all the samples meet the contract specifications
D) fluctuation ( unstable Asphalt content )
Explanation:
B) The daily average asphalt content has to obtained in order to determine the upper and lower control limits using an average asphalt content of 5.5% +/- 0.5% everyday
The upper limit : 14 may = ( 5.8 + 5.1 ) / 2 = 5.85
The lower limit : 16 may = ( 5.2 + 4.8 ) / 2 = 5.0
attached below is the required plot
C ) Not all the samples meet the contract specifications and the samples that do not meet up are samples from :
15 may and 16 may . this is because their Asphalt contents are 6.2 and 4.8 respectively and sample number 18 and 20
D ) what can be observed is that the ASPHALT content fluctuates between the dates while the contract specification is fixed
Answer:
3.03 INCHES
Explanation:
According to ASTM D198 ;
Modulus of rupture = ( M / I ) * y ----- ( 1 )
M ( bending moment ) = R * length of span / 2
= (120 * 10^3 ) * 48 / 2 = 288 * 10^4 Ib-in
I ( moment of inertia ) = bd^3 / 12
= ( 2 )*( d )^3 / 12 = 2d^3 / 12
b = 2 in , d = ?
length of span = 4 * 12 = 48 inches
R = P / 2 = 240 * 10^3 / 2 = 120 * 10^3 Ib
y ( centroid distance ) = d / 2 inches
back to equation ( 1 )
( M / I ) * y
940.3 ksi = ( 288 * 10^4 / 2d^3 / 12 ) * d / 2
= ( 288 * 10^4 * 12 ) / 2d^3 ) * d / 2
940300 = 34560000* d / 4d^3
4d^3 ( 940300 ) = 34560000 d ( divide both sides with d )
4d^2 = 34560000 / 940300
d^2 = 9.188 ∴ Value of d ≈ 3.03 in
Answer:
(a) Flow rate of vehicles = No of vehicles per mile * Speed
=No of cars per mile * Speed +No of trucks per mile * Speed
= 0.75*50*60 + 0.25*50*40
=2750 vehicles / hour
(b) Let Density of vehicles on grade = x
Density on flat * Speed =Density on grade * Speed
So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25
So, x= 57.89
So, Density is around 58 Vehicles per Mile.
(c) Percentage of truck by aerial photo = 25%
(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %
Answer: I am not for sure
Explanation:
