To solve this problem it is necessary to apply the concepts related to rate of thermal conduction
The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, 
, T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.
The change made between glass and air would be determined by:
There are two layers of Glass and one layer of Air so the total temperature would be given as,
Finally the rate of heat flow through this windows is given as,
Therefore the correct answer is D. 180W.
Answer:
C.) Wave B has about 2.25 times more energy than wave A
Explanation:
<em>500 sec</em>
<em>8 min 20 sec</em>
<em>Hi there !</em>
<em />
<em>8 m ................ 1 s </em>
<em>4000 m ........ x s</em>
<em>x = 4000m×1s/8m = 500 sec = 8 min 20 sec</em>
<em />
<em>Good luck ! </em>
Answer:
Isabella will not be able to spray Ferdinand.
Explanation:
We'll begin by calculating the time taken for the water to get to the ground from the hose held at 1 m above the ground. This can be obtained as follow:
Height (h) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =.?
h = ½gt²
1 = ½ × 9.8 × t²
1 = 4.9 × t²
Divide both side by 4.9
t² = 1/4.9
Take the square root of both side
t = √(1/4.9)
t = 0.45 s
Next, we shall determine the horizontal distance travelled by the water. This can be obtained as follow:
Horizontal velocity (u) = 3.5 m/s
Time (t) = 0.45 s
Horizontal distance (s) =?
s = ut
s = 3.5 × 0.45
s = 1.58 m
Finally, we shall compare the distance travelled by the water and the position to which Ferdinand is located to see if they are the same or not. This is illustrated below:
Ferdinand's position = 10 m
Distance travelled by the water = 1.58 m
From the above, we can see that the position of the water (i.e 1.58 m) and that of Ferdinand (i.e 10 m) are not the same. Thus, Isabella will not be able to spray Ferdinand.
Answer:
The object starts away from the origin and then moves toward the origin at a constant velocity. Next, it stops for one second. Finally, it moves away from the origin at a greater constant velocity.
