Answer:
1/7
Step-by-step explanation:
&&5_4_&&&&&&&&&&&&&---
<span>1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
Formula: Heat = mass * specific heat * ΔT
Data:
mass = 50 g = 0.050 kg
specific heat of iron = 450 J/ kg °C
ΔT = 124°C - 10°C ¿ 114 °C
=> heat = 0.050kg * 450 J / kg°C * 114°C ≈ 2.6 J
2. If a refrigerator is a heat pump that follows the first law of
thermodynamics, how much heat was removed from food inside of the
refrigerator if it released 492J of energy to the room? Joules
The firs law of thermodynamics is conservation of energy => energy removed from inside of the refrigerator = energy released to the room
=> Answer = 492 J
3. How much heat is needed to raise the temperature of 45g of water by 63oC? Joules
Formula: heat = mass * specific heat * ΔT
specific heat of water = 4186 J / Kg °C
heat = 0.045 kg * 4186 J/kg °C * 63°C = 11,867.31 J
</span>
Answer:
h=12 A=144 sq. cm.
Step-by-step explanation:
pythagoream therom:
9^2+h^2=15^2
81+h^2=225
h^2=225-81
h^2=144
Sq. rt h^2=sq. rt 144
h=12cm
1/2(b1+b2)h
1/2(5+19)12
1/2(24)12
1/2(288)
a=144
Answer:
14 millimeters
Step-by-step explanation:
V=(4/3)πr³
1372π(1/3)=(4/3)πr³
1372=4r³ <em>(cancel out the pi and multiply both sides by 3)</em>
r³=343
r=7
Diameter is twice the radius = 2(7) = 14
Answer: ![\frac{\sqrt[4]{10xy^3}}{2y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D)
where y is positive.
The 2y in the denominator is not inside the fourth root
==================================================
Work Shown:
![\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%7D%7B8y%7D%7D%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%2A2y%5E3%7D%7B8y%2A2y%5E3%7D%7D%5C%20%5C%20%5Ctext%7B....%20multiply%20top%20and%20bottom%20by%20%7D%202y%5E3%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B10xy%5E3%7D%7B16y%5E4%7D%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B16y%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20break%20up%20the%20fourth%20root%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20rewrite%20%7D%2016y%5E4%20%5Ctext%7B%20as%20%7D%20%282y%29%5E4%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D%20%5C%20%5C%20%5Ctext%7B...%20where%20y%20is%20positive%7D%5C%5C%5C%5C%5C%5C)
The idea is to get something of the form 
in the denominator. In this case, 
To be able to reach the 
, your teacher gave the hint to multiply top and bottom by 
For more examples, search out "rationalizing the denominator".
Keep in mind that ![\sqrt[4]{(2y)^4} = 2y](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%202y)
only works if y isn't negative.
If y could be negative, then we'd have to say ![\sqrt[4]{(2y)^4} = |2y|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%20%7C2y%7C)
. The absolute value bars ensure the result is never negative.
Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.
