Heat Transfer by conduction in solids, convection of fluids (liquids or gases), and radiation through anything that will allow radiation to pass. if this is not what are you looking for I am sorry :c
Answer:
atomic concentration = 2 atoms/unit cell
lattice parameter: a= 3.22 x 10⁻¹⁰ m
atomic radius: r= 1.39 x 10⁻¹⁰m
Explanation:
The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell
In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:
Vc= a³= 
a³= 3.12 x 10⁻²³ m³
⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m
Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:
a= 
⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m
D. The equation for Velocity is v=d/t. Given the distance and time spent, velocity can be determined.
Answer:
2a) x = 32 [mil/h]; 2b) t = 0.5[h]; 3a) t = 2.5 [h]; 3b) x = 185[mil]
Explanation:
2a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 32 [\frac{mil}{h}] \\t=time = 1 [h]\\x=v*t\\x=32[\frac{mil}{h} ]*1[h]\\x=32[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2032%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%201%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D32%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A1%5Bh%5D%5C%5Cx%3D32%5Bmil%7D)
2b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{420}{840}\\ t=0.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B420%7D%7B840%7D%5C%5C%20t%3D0.5%5Bh%5D)
3a)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\t=\frac{x}{v} \\t=\frac{35}{14}\\ t=2.5[h]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Ct%3D%5Cfrac%7Bx%7D%7Bv%7D%20%5C%5Ct%3D%5Cfrac%7B35%7D%7B14%7D%5C%5C%20t%3D2.5%5Bh%5D)
3b)
We can solve this problem by using the kinematics equation, which relates speed to time and displacement.
![v=\frac{x}{t} \\v=velocity [\frac{mil}{h} ] = 74 [\frac{mil}{h}] \\t=time = 2.5 [h]\\x=v*t\\x=74[\frac{mil}{h} ]*2.5[h]\\x=185[mil}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5Cv%3Dvelocity%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%20%3D%2074%20%5B%5Cfrac%7Bmil%7D%7Bh%7D%5D%20%5C%5Ct%3Dtime%20%3D%202.5%20%5Bh%5D%5C%5Cx%3Dv%2At%5C%5Cx%3D74%5B%5Cfrac%7Bmil%7D%7Bh%7D%20%5D%2A2.5%5Bh%5D%5C%5Cx%3D185%5Bmil%7D)
Answer:
Induced EMF in the loop is 80 volts.
Explanation:
Given that,
Magnetic field, 
Time, 
Area of the loop, 
We need to find the magnitude of the average value of the emf induced in the loop. Due to change in magnetic field, an emf is induced in the loop which is given by :




So, the value of induced emf is 80 V. Hence, this is the required solution.