b. Mars
c. the Moon
d. Polaris
What is the distance an object would be from Earth if its parallax were one arcsecond?
a. a parsec
b. a light-year
c. an arcminute
d. an angular diameter
A) The temperature at the fat-inner fur boundary be so that the bear loses heat at a rate of 51.4W is; T_i = 38.52°C
B) The thickness of the layer contained within the fur so that the bear loses heat at a rate of 51.4 W is; t = 13.41 cm
We are given;
Diameter of sphere; d = 1.6 m
Radius of sphere; r = d/2
r = 1.6/2
r = 0.8 m
Thickness of bear; t = 3.9 cm cm = 0.039 m
Outer surface Temperature of fur; T_h = 2.8 ∘C
Inner surface Temperature of fat;T_f = 30.9 ∘C
Thermal conductivity of fat; K_f = 0.2 W/m⋅k
Thermal conductivity of air; K_a = 0.024 W/m⋅k
A) To find the temperature at the fat-inner fur boundary when heat loss is 51.4 W, we will use the heat current formula;
H = K_f•A(T_f - T_i)/t
Where;
A is area = 4πr²
A = 4π × 0.8²
A = 8.04 m²
T_i is the temperature we are looking for
H is heat loss = 51.4
t is thickness
Making T_i the subject gives;
T_i = (T_f × H × t)/(K_f × A)
T_i = (30.9 × 51.4 × 0.039)/(0.2 × 8.04)
T_i = 38.52°C
B) We want to find the thickness of the layer contained within the fur. Thus, we will use K_a instead of K_f. Let us make t the subject in the heat current formula to get;
t = (K_a•A(T_i - T_h)/H
t = (0.024 × 8.04 × (38.52 - 2.8))/51.4
t = 0.1341 m
t = 13.41 cm
Read more at; brainly.com/question/14548124
Answer:
Explanation:
Using the equation
H = Q/t = k A ( T hot - T cold) / L
where H is the rate of heat loss = 51.4 W, T cold be temperature of the outer surface, A is the surface area of the fat layer which is a model of sphere ( surface area of a sphere ) = 4πr² where diameter = 1.60 m
radius = 1.60 m / 2 = 0.80 m
A = 4 × 3.142 × ( 0.8²) = 8.04352 m²
making T cold subject of the formula
T cold = T hot - = 30.9° C - ( 51.4 W × 3.9 × 10⁻² m) / ( 0.2 W/mK × 8.04352 m² ) = 30.9° C - 1.25 ° C = 29.65° C
b) The thickness of air layer for the bear to lose heat t a rate of 51.4 W
thermal conductivity of air is 0.024 W/mK and rearranging the earlier formula
L = = (0.024 W/ m K × 8.04352 m²) ( 29.65° C - 2.8°C) / 51.4 W = 0.101 m = 10.1 m
Physical properties are properties that you can see, feel and measure, they do not change the matter of the substance. Some examples of physical properties are Mass, Color, Shape, Hardness and Boiling point.
When two electron in a chemical bond are equally shared between two different atoms, the bond is covalent bond.
A chemical bond known as a covalent bond includes the exchanging of electrons between atoms to create electron pairs.
These electron couples are referred to as bonding pairs or sharing pairs. Covalent bonding is the stable equilibrium of attracting and repulsive forces that exists when two atoms share an electron.
The ability of each atom to reach the equivalent of a full valence shell, which corresponds to a stable electronic state, is made possible for many molecules by the sharing of electrons. Hence, when two electron in a chemical bond are equally shared between two different atoms, the bond is covalent bond.
Learn more about covalent bond here:
#SPJ6