Answer:
576.21kJ
Explanation:
#We know that:
The balance mass
so,
#Also, given the properties of water as;
#We assume constant properties for the steam at average temperatures:
#Replace known values in the equation above;
#Using the mass and energy balance relations;
#We have : we replace the known values in the equation as;
#Hence,the amount of heat transferred when the steam temperature reaches 500°C is 576.21kJ
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:
Yes, fracture will occur
Explanation:
Half length of internal crack will be 4mm/2=2mm=0.002m
To find the dimensionless parameter, we use critical stress crack propagation equation
and making Y the subject
Where Y is the dimensionless parameter, a is half length of crack, K is plane strain fracture toughness, is critical stress required for initiating crack propagation. Substituting the figures given in question we obtain
When the maximum internal crack length is 6mm, half the length of internal crack is 6mm/2=3mm=0.003m
and making K the subject
and substituting 260 MPa for while a is taken as 0.003m and Y is already known
Therefore, fracture toughness at critical stress when maximum internal crack is 6mm is 42.455 Mpa and since it’s greater than 40 Mpa, fracture occurs to the material
Answer:
The Current will decrease by a factor of 2
Explanation:
Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.
Now, as the question says, the load is reduced to half its original value, we can write:
Since, P2 = P1/2,
Dividing equations (1) and (2), we get,
P1 / (P1/2) = I1/ I2
Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.