Answer:
9500 kJ; 9000 Btu
Explanation:
Data:
m = 100 lb
T₁ = 25 °C
T₂ = 75 °C
Calculations:
1. Energy in kilojoules
ΔT = 75 °C - 25 °C = 50 °C = 50 K

2. Energy in British thermal units

The equations are based on the following assumptions
1) The bar is straight and of uniform section
2) The material of the bar is has uniform properties.
3) The only loading is the applied torque which is applied normal to the axis of the bar.
4) The bar is stressed within its elastic limit.
Nomenclature
T = torque (Nm)
l = length of bar (m)
J = Polar moment of inertia.(Circular Sections) ( m^4)
J' = Polar moment of inertia.(Non circluar sections) ( m^4 )
K = Factor replacing J for non-circular sections.( m^4)
r = radial distance of point from center of section (m)
ro = radius of section OD (m)
τ = shear stress (N/m^2)
G Modulus of rigidity (N/m^2)
θ = angle of twist (radians)
Answer:
A)
It should be Non- toxic
It should possess high Thermal conductivity
It should have the Required Thermal diffusivity
B)
- stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
- porcelain: mostly used for mugs and it is non-toxic
- Pyrex : posses good thermal conductivity used in oven
C) All the materials are suitable because they serve different purposes when making modern kitchen cookware
Explanation:
A) characteristics required of a ceramic material to be used as a kitchen cookware
- It should be Non- toxic
- It should possess high Thermal conductivity
- It should have the Required Thermal diffusivity
B) comparison of three ceramic materials as to their relative properties
- stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
- porcelain: mostly used for mugs and it is non-toxic
- Pyrex : posses good thermal conductivity used in ovens
C) material most suitable for the cookware.
All the materials are suitable because they serve different purposes when making modern kitchen cookware
Answer:

Explanation:
Density is defined as mass ler unit volume, expressed as

Where m is mass,
is density and v is the volume. For a sphere, volume is given as

Replacing this into the formula of density then

Given diameter of 2.24 in then the radius is 1.12 in. Substituting 0.82 lb for m then

Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂