Answer:
1.312 x 10⁻¹² J/nucleon
Explanation:
mass of ¹³⁶Ba = 135.905 amu
¹³⁶Ba contain 56 proton and 80 neutron
mass of proton = 1.00728 amu
mass of neutron = 1.00867 amu
mass of ¹³⁶Ba = 56 x 1.00728 amu + 80 x 1.00867 amu
= 137.10128 amu
mass defect = 137.10128 - 135.905
= 1.19628 amu
mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg
= 1.9858 x 10⁻²⁷ Kg
speed of light = 3 x 10⁸ m/s
binding energy,
E = mass defect x c²
E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²
E = 17.87 x 10⁻¹¹ J/atom
now,
binding energy per nucleon =
= 0.1312 x 10⁻¹¹ J/nucleon
= 1.312 x 10⁻¹² J/nucleon
Answer:
C. Approx. Two thirds of the water is in the ICF compartment
Explanation:
The body cells are bathed in fluids internally and externally. The water inside the cells make up about 42% of the total body weight and is called the intracellular fluid (ICF). The rest of the fluid outside the cells is called extracellular fluid (ECF) and is separated from the intracellular fluid by a semipermeable membrane that surrounds the cell, and only allows fluid to flow in and out of the cells, but prevents unwanted molecules or materials from getting in.
All I know is that the more proton and electrons an atom has the less metallic it becomes. But other than that I think it's pretty random
Answer:
40.25 V
Explanation:
Parameters given:
Initial magnetic field, B1 = 0.65 T
Final magnetic field, B2 = 0 T
Number of turns, N = 73
Radius of loop, r = 18 cm = 0.18 m
Time taken, Δt = 0.12 s
The magnitude of the EMF induced in a loop of wire due to a changing magnetic field is given as:
V = | [-N * A * (B2 - B1)] / Δt |
Where A = area of loop
Area of loop = πr²
Therefore, the magnitude of the EMF induced will be:
V = | [-N * πr² * (B2 - B1)] / Δt |
V = | [-73 * π * 0.18² * (0 - 0.65)] / 0.12 |
V = |40.25| V
V = 40.25 V
This is the magnitude of the EMF induced in the loop of wire.
