Answer:
a)The electric Field will be zero at the point between the sheets
b)
c)
Explanation:
Let 
be the surface charge density of the of the non conducting parallel sheet.Let consider a Gaussian surface in the form of of cylinder such that its cross-sectional is A . Then there will be flux only due to cross sectional area as the curved sectional is perpendicular to the the electric field so the Electric Flux due to it is zero.
Now using Gauss law we have, E be the electric Field at the distance r from the sheet then
The Field will be away from the sheet and perpendicular to it.
a) The Electric Field between them
b)The Electric Field to the right of the sheets
c)The Electric Field to the left of the sheets
Answer:
8.60 g/cm³
Explanation:
In the lattice structure of iron, there are two atoms per unit cell. So:
where 
an and A is the atomic mass of iron.
Therefore:
This implies that:
= 
Assuming that there is no phase change gives:
= 8.60 g/m³
Answer:
4 N
Explanation:
mass = 2 kg
acceleration = 2 m/s^2
Force = mass * acceleration
= 2 *2
= 4 N
Answer: Dependent Unit or System of Units
Explanation:
Density is calculated by dividing mass (Kg) by volume (L).
The unit of Density is Kg/L or one of their derivatives such as g/cm³.
Explanation:
12) q = mCΔT
125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)
T = 82.0°C
13) Solving for ΔT:
ΔT = q / (mC)
a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C
b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C
c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C
d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C
e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C
f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C
14) q = mCΔT
q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)
q = 502,000 J
20) q = mCΔT
q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)
q = 742,000 J
24) q = mCΔT
q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)
q = -0.091 J
