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4. For this problem, we have to write and solve a proportion. We would set this proportion up as 12/15 = 8/x. This is because we're looking for the length of the shadow and we know the height of the items, so we line them up horizontally and x goes with 8, because we're looking for the shadow length. Let's cross multiply the values. 15 * 8 = 120. 12 * x = 12. You get 120 = 12x. Now, we must divide each side by 12 to isolate the "x". 120/12 is 10. x = 10. There. The cardboard box casts a shadow that is 10 ft long.
5. For this question, you do the same thing. This time, you're finding the height of the tower, so you would do 1.2/0.6 = x/7. Cross multiply the values in order to get 8.4 = 0.6x. Now, divide each side by 0.6x to isolate the "x". 8.4/0.6 is 14. x = 14. There. The tower is 14 m tall.
If you need more help on proportions and using proportions in real life situations, feel free to search on the internet to find more information about how you solve them.
Answer:
a
b
Explanation:
From the question we are told that
The pressure of the water in the pipe is 
The speed of the water is 
The original area of the pipe is 
The new area of the pipe is ![A_2 = \pi * \frac{[\frac{d}{2} ]^2}{4} = \pi * \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}](https://tex.z-dn.net/?f=A_2%20%3D%20%5Cpi%20%2A%20%20%5Cfrac%7B%5B%5Cfrac%7Bd%7D%7B2%7D%20%5D%5E2%7D%7B4%7D%20%20%3D%20%20%5Cpi%20%2A%20%20%5Cfrac%7B%5Cfrac%7Bd%5E2%7D%7B4%7D%20%7D%7B4%7D%20%3D%20%5Cpi%20%5Cfrac%7Bd%5E2%7D%7B16%7D)
Generally the continuity equation is mathematically represented as
Here 
is the new velocity
So
=> 
=> 
=> 
=>
Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the Bernoulli's equation for constant height to calculate the pressure
This is mathematically represented as
Here 
is the density of water with value 
![P_2 = P_1 + \frac{1}{2} * \rho [ v_1^2 - v_2^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20P_1%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%20%5Crho%20%5B%20v_1%5E2%20-%20v_2%5E2%20%5D)
=> ![P_2 = 110 *10^{3} + \frac{1}{2} * 1000 * [ 1.4 ^2 - 5.6 ^2 ]](https://tex.z-dn.net/?f=P_2%20%3D%20%20110%20%2A10%5E%7B3%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20%201000%20%2A%20%20%5B%201.4%20%5E2%20-%205.6%20%5E2%20%5D)
=>
and C. corrosive, increases the concentration of hydrogen ions when added to water, forms hydrogen gas when it comes in contact with a metal, and formssalt and water when added to a base.
First, balance the reaction:
_ KClO₃ ==> _ KCl + _ O₂
As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :
2 KClO₃ ==> 2 KCl + 3 O₂
Since we start with a known quantity of O₂, let's divide each coefficient by 3.
2/3 KClO₃ ==> 2/3 KCl + O₂
Next, look up the molar masses of each element involved:
• K: 39.0983 g/mol
• Cl: 35.453 g/mol
• O: 15.999 g/mol
Convert 10 g of O₂ to moles:
(10 g) / (31.998 g/mol) ≈ 0.31252 mol
The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need
(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃
KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of
(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g
of KClO₃.
