Was D the correct answer?
15.0cm
45.0cm
55.0cm
60.0cm
85.0cm
Answer:
Explanation:
its 85 cm because it just is in meters it would be .8m which would be decent distance to hear it
B. rises toward the ceiling.
C. transfers heat to the walls.
D. sinks toward the floor.
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
this set of data? Which axis of a line graph does it normally go on?
b. thermal energy
c. mechanical energy
d. electrical energy
Answer:
Therefore, a. chemical energy plant store when light is transformed during photosynthesis.
Explanation:
Photosynthesis is the process plants and some algae use to convert light energy to chemical energy stored as sugar within chloroplasts .
Photosynthesis is 6CO2 + 6H20 + light energy = glucose + 6 oxygen.
Photosynthesis is divided into two main stages: light reaction and dark reaction. The light reaction converts light energy into adenosine triphosphate,
Potential energy or stored energy, and kinetic energy, the energy due to motion can be balanced in the process of converting kinetic energy to potential energy during an uphill motion
The correct option for, which situation shows potential energy and kinetic energy are balanced is option;
A roller coaster car going uphill
The reason the selected option is correct is as follows:
Potential energy is the energy that is due to the relative position of an item in relation to a ground or zero state. The formula for potential energy due to the elevation is given as follows;
Potential energy, P.E. = m·g·h
Kinetic energy is the energy that is due to motion. The kinetic energy of an item is given as follows;
Kinetic energy, K.E. = (1/2) × m × v²
The potential and kinetic energy of a body is balanced when we have;
P.E. = m·g·h = K.E. = (1/2)·m·v²
Which gives;
g·h = (1/2)·v²
Therefore, a point is reached as the an body moves up a heal, where the potential energy (the energy due to height of the object) and the kinetic energy (the energy due to current speed) of the object are equal
The correct situation which shows potential energy and kinetic energy are balanced is therefore; A roller coaster car going uphill
Learn more about potential and kinetic energy here: