The journal entry to record this transaction consists of a: debit of the purchases account for $21400 and credit the accounts payable account for $21400
<h3>How to enter a transaction in a journal?</h3>
We are told that a company purchased $21,400 of supplies on credit.
Now, it is vital to note that in a journal, there is the credit column and also the debit column.
Now, the way we carry out recording the Journal Entry for a Credit Purchase is that;
When a business purchases goods or services on credit, the business will then debit the purchases account, which will increase the business’s assets.
The business will also credit the accounts payable account, which will increase the business’s liabilities.
Thus;
The journal entry to record this transaction consists of a: debit of the purchases account for $21400 and credit the accounts payable account for $21400
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Answer:
X has travelled 4 times as far as Y
X is moving 2 times as fast as Y
Explanation:
Because x=1/2at²(get rid of the first portion of x=vt+1/2at² because initial v is 0 which multiplies to 0). If the acceleration is set to for example 10 (keep this constant for both X and Y) and we use t=1 and t=2 for time (as the time travelled of X is twice that of Y). Plugging this into the equation, at t=1, x=4. At t=2, x=20. 20 is four times greater than 5, so X has travelled 4 times as far as Y.
To find the difference in speed between the two objects, use the equation 
. Since the initial velocity is 0, that part can just be removed from the equation. With v=at, it is easy to see that if the time plugged in is twice for one than the other (and the acceleration is the same for both), the final result will be twice of the other as well. For example: If the acceleration is 10 again for both, then v=10t. If t is 1, the velocity is 10. If t is 2, the velocity is 20.
1) B.Metallic windows
2) D. Absorption
3) A. Clear glass
The mass of glucose solute dissolved in the solution is 6.739 Kg.
Recall that;
ΔT = K m i
ΔT = Freezing point depression
K =Freezing point depression constant = 1.86°C/mol
m = molality of the solution
i = Van't Hoff factor = 1 (molecular solution)
We have to find the freezing point depression from;
Freezing point depression = Freezing point of pure solvent - Freezing point of solution
Freezing point of pure water = 0°C
Freezing point of solution = -5. 8 ∘C
Freezing point depression = 0°C - (-5. 8 ∘C) = 5. 8 ∘C
Now;
m = ΔT/K i
m = 5. 8 ∘C/ 1.86°C/mol × 1
m = 3.12 m
But molality = number of moles of solute/mass of solvent in Kg
Molar mass of solute = 180 g/mol
Let the mass of solute be m
3.12 = m/180/12
3.12 = m/180 × 12
m = 3.12 × 180 × 12
m = 6739 g or 6.739 Kg
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