Beaker A contains more water.
First,find equivalent fraction.
Next,compare the amounts you will see Beaker A contains more water.
Therefore,Beaker A contains more water.
Answer:
-45
Step-by-step explanation:
-34+15=-19
-29-(-3)= -29+3= -26
-19 + -26 = -45
Answer:
0.0003W/cm°C
Step-by-step explanation:
The question is not properly written. Here is the correct question.
The batting wang xiu ying uses to fill quilts has a thermal conductivity rate of 0.03 watts (W) per meter(m) per degree celsius. what is the batting thermal conductivity when w/cm•c
Given the thermal conductivity in W/m°C to be 0.03W/m°C
We are to rewrite the value in W/cm°C
The difference is the unit. The only thing we need to do is to simply convert the unit (metres) in W/m°C to centimeters (cm)
Since 100cm = 1m, 0.03W/m°C can be expressed as shown below;
= 0.03W/m°C
= 0.03 × W/1m×°C
Note that 1m = 100cm, substituting this conversion into the expression, it will become;
= 0.03 × W/100cm × °C
= 0.03/100 × W/cm°C
= 0.0003W/cm°C
Hence the battling thermal conductivity in W/cm°C is 0.0003W/cm°C
Answer:
The answer is C) y - 2 = -5/7(x - 6)
Step-by-step explanation:
Since we have the slope (-5/7) and a point (6, 2), we can just input those in for m and (x1, y1) in point-slope form.
y - y1 = m(x - x1)
y - 2 = -5/7(x - 6)
Step 1:
<span>Calculate the effective thermal conductivity of the wall or ceiling:
</span>
K_eff = [ (13 ÷ 8)(0.12) + (16 - (13 ÷ 8)) × (0.04)] ÷ 16
K_eff =<span> [ 0.195 + 0.565] </span>÷<span> 16
</span>
K_eff = 0.76 ÷ 16
K_eff = 0.0475 W/ (m K)
Step 2:
Calculate <span>the interior ceiling area:
</span>Area of each of the interior side walls = <span>8.82 m x 8.64 m
= 76.2 m</span>²
Area of the interior ceiling = 8.64 m × <span>8.64 m
</span> = 74.6 m²
H = - k·A·(Δ - T) ÷ <span>(thickness)
</span>
H = - 0.0475 ÷ (379.45 × 20) ÷ 45/8
H = - ( - 0.95 × 379.45 ) ÷<span> 0.1429
</span>
H = <span>2.52 kW </span>
