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taylor perez

taylor p.

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college eduka Home Gelen College X Q2 Xespan Development X Topic Growth and De X prs/classroom.galencollege.edu/courses/7519740/external tools/retrieve display full width url-https%3A%2F%2Fgalencollege.quiz-t-ad-prod 3 2 Multiple Choice 10 points 11:09 Time Remaining < The process which we obtain energy from fat, triglycerides must first be broken down by hydrolysis into fatty acids and glycerol is Oglycolysis Οpolysis lipogenesis Oglucogenesis 123 Q Search 4 5 6 7 8 9 Q W A S ERT DEGH P J K L < Z X CVB N M alt

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Which of the following statements concerning a voltaic cell is/are correct? 1. Oxidation occurs at an anode. [Select] A spontaneous reaction generates an electric current in a voltaic cell. 2. [Select] Without a salt bridge, charge buildup will cause the cell reaction to stop. 3. [Select]

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A strain of mold was grown in a batch culture and the following data were obtained. Time (hours) Cell Concentration (g/L) 0 1.25 9 2.45 16 5.1 23 10.5 30 22 34 33 35 37.5 40 41

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Question 2 Not yet answered Marked out of 1.00 The lungs, nose, and trachea are part of which organ system? O a. Respiratory O b. Digestive O c. Endocrine Previous page

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Clicks Pte Ltd is an e-commerce company that offers online shopping in Singapore. Following is the extract of the unadjusted trial balance of Clicks Pte Ltd as at 31 December 2021. Account title Debit ($) Credit ($) Share capital (100,000 shares at $1.00 each) 500,000 Retained earnings, 31 Dec 2020 110,000 Freehold property at cost 1,000,000 Motor Vehicle at cost 80,000 Cost of goods sold 2,254,800 Sales 3,129,200 Wages and salaries 480,000 Allowance for doubtful debts 6,000 Accounts receivable 96,800 Accounts payable 233,200 Bank 40,000 Bank Loan (for delivery van) 64,000 Inventory 88,400 Rental income 60,000 Interest expense 1,600 Additional information: (i) The company bought a delivery van on 1 January 2021 that cost $80,000. The delivery van is expected to have a residual value of $8,000 at the end of its useful life. This purchase has been recorded in the accounts. (ii) As the company does not occupy the entire freehold property, it rented space to two tenants at $5,000 per month each starting on 1 July 2021. The first tenant paid $15,000 for three months’ rent on 1 July 2021. However, no further payment has been made by this tenant. At the same time, on 1 July 2021, second tenant paid $45,000 for nine months’ rental. All payments received have been recorded as rental income. (iii) On 1 January 2021, to finance the purchase of the delivery van, the company took a loan of $64,000 from the bank. The company only needs to start making the first principal repayment on 1 January 2022. The bank charged an interest of 5% per annum. Interest for the loan is payable on 1 January and 1 July. Interest for the six months had been paid on 1 July 2021. This has also been recorded in the accounts. (iv) Wages and salaries due but still remained unpaid at 31 December 2021, $7,550 (v) The company estimated that as at 31 December 2021, 5% of accounts receivable will be uncollectible. (vi) No depreciation has been charged for the year ended 31 December 2021. The company depreciates motor vehicle over a useful life of five years using the double declining balance method. Required: (a) Compute the necessary 31 December 2021 adjusting journal entries. No narrations required.

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(b) Calculate the heat required to bring 300 kg of solid sodium from 0°C to a vapour at 1000°C at constant pressure of 1 atm. Latent heat of fusion of sodium at 97.7°C = 2.636 kJ mol$^{-1}$ Latent heat of vaporization of sodium at 914°C = 96.734 kJ mol$^{-1}$ Heat capacity of liquid sodium = 31.38 J mol$^{-1}$ K$^{-1}$ Heat capacity of sodium vapour = 20.79 J mol$^{-1}$ K$^{-1}$ Heat capacity of solid sodium = (20.96 + 0.0224T) J mol$^{-1}$ K$^{-1}$ where T is in K Molecular weight of sodium = 23 (15 marks)

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Exercise 2: If the voltage $v_o$ across the $2\Omega$ resistor in the circuit presented by Figure on the right is $10cos2t$ V, obtain $i_s$. $0.1 F \longrightarrow \frac{1}{j\omega C} = \frac{1}{j(2)(0.1)} = -j5$ $0.5 H \longrightarrow j\omega L = j(2)(0.5) = j$ The current I through the 2-$\Omega$ resistor is $I = \frac{1}{1 - j5 + j + 2} = \frac{I_s}{3 - j4}$ $I_s = (5)(3 - j4) = 25\angle -53.13^\circ$ where $I = \frac{10}{2}\angle 0^\circ = 5$ Therefore, $i_s(t) = 25 \cos(2t - 53.13^\circ) A$

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3. (20 points) For some transformation at temperature 360K, its kinetics obeys the Avrami equation, the parameter $n$ is known to have a value of 1.5. If the reaction is 25\% complete after 90s, what would be the percent of transformation when the reaction continue to take 270s?

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B, Priority=32768, MAC=0000.2222.3333 A, Priority=100, MAC=0000.1111.2222 D, priority=200, MAC=0000.4444.5555 C, Priority=32768, MAC=0000.6666.7777 E, Priority=32768, MAC=0000.8888.9999 Q2: Choose the root ports and designated ports in this example. - All edges have 1Gbps speed at cost=4 - In case of choocing a blocking port write down the reason along.

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1.1 The Compton scattering is depicted in the figure below. Incident photon $E_\gamma$ $E_\gamma'$ scattered photon $\theta$ $\phi$ scattered electron Using the above figure with its symbols and the notion of relativistic dynamics, derive (explaining each step) the Compton-scattering formula: $\frac{E_\gamma'}{1 + (E_\gamma/mc^2)(1-\cos\theta)} = E_\gamma$

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