Answer:
So the fraction of proeutectoid cementite is 44.3%
Explanation:
Given that
cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.
We know that
Fraction of proeutectoid cementite phase gievn as

Now by putting the values


So the fraction of proeutectoid cementite is 44.3%
Answer:
The density of the copper is higher than aluminium. Hence it is heavier compared to aluminium conductors it requires strong structures and hardware to bear the weight. More ductile and has high tensile strength.
...
Aluminium & Copper properties.
Property Copper (Cu) Aluminium (Al)
Density (g/cm3) 8.96 2.70
The three most common software development methods are the Waterfall Approach, the Incremental Approach, and the SPIRAL Approach. These methods depend on the team size and specific goals.
Software development is the sequential procedure that involves the division of the work into smaller and parallel stages in order to improve software design and product management.
The software development methods depend on both the team size and specific objectives.
The most common methodologies for software development include:
- Waterfall
- Spiral
- Incremental
- Agile
- Continuous integration
Learn more about software development here:
brainly.com/question/14275830
Answer:
Isn't the answer written upside down in the sentence?
Explanation:
Answer:
W = 1 mJ
V_new = 2000 V
W_new = 2 mJ
The extra energy came from the work done from moving the plates
Explanation:
We are given;
Capacitance; C = 1000pF = 10^(-9) F
Voltage; V = 1000V
Now,formula for stored energy in a parallel plate capacitor is given by;
W = (1/2)CV² = (1/2)(10^(-9))(1000²) = 0.5 mJ
However, in this case, it's W = 0.5 x 2 = 1 mJ since parallel-plate capacitor with air dielectric
When plates are moved and distance between plates is doubled(net charge is the same), thus we can calculate voltage from;
Q = CV
Since, C_new = C/2
Thus,
Q = (C/2)V_new
V_new = 2Q/C
Thus, V_new = 2V
Thus, V_new = 2 x 1000 = 2000 V
Now,
W_new = (1/2)C_new•(V_new)² = (1/2) (0.5C)•(2V)² = CV² = 2W = 2 x 1 = 2mJ