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- $[a] = \frac{P \times V^{2}}{n^{2}}$
- In common units, this translates to $\text{atm} \times (\text{dm}^{3})^{2} / \text{mol}^{2}$
- Result: $atm \text{ dm}^{6} \text{ mol}^{-2}$
As part of their sports physical, seven college athletes - F, G, H, I, J, K and L - are being weighed. In announcing the results of the physical exams, the coach has given the following information.
i . None of the athletes is exactly the same weight as another athlete.
ii. K is heavier than L, but lighter than H.
iii. I is heavier than J
iv. Both F and G are heavier than H.
Sub-Questions:
The textbook discusses noise in the Shannon-Weaver model context and identifies three categories:
- Physical (Mechanical/Engineering) Noise: Unexplained variation or random error in the communication channel. Examples: a loud motorbike during conversation, smudges on a printed page, “snow” on a TV set, mist on a car windscreen, a crackling microphone. Shannon was primarily concerned with this type.
- Semantic Noise: Arises from differences in meaning — when sender and receiver do not share the same knowledge, cultural background, experience, attitudes, beliefs, or linguistic skills. This is “the very essence of the study of human communication.”
- Psychological Noise: The textbook mentions “psychological interference with encoding and decoding” — arising from emotional states, biases, prejudices, or mental conditions that distort how messages are encoded or decoded.
Channel overload — when the channel capacity is exceeded — is also mentioned as a source of distortion distinct from noise per se.
Using the superposition principle: Potential at cavity centre = (Potential due to full sphere at that point) $-$ (Potential due to removed small sphere at its own centre).
The cavity centre is at distance $R/2$ from the big sphere centre.
Potential due to full sphere at $r = R/2$ (inside):
$V_{full} = -\frac{GM}{2R^3}\left(3R^2 - \frac{R^2}{4}\right) = -\frac{11GM}{8R}$
Potential due to removed sphere (mass $M/8$, radius $R/2$) at its own centre:
$V_{small} = -\frac{3G(M/8)}{2(R/2)} = -\frac{3GM}{8R}$
$V_{cavity} = V_{full} - V_{small} = -\frac{11GM}{8R} + \frac{3GM}{8R} = -\frac{GM}{R}$
At the end of a financial period, a business recorded: (1) Non-current assets at carrying value, (2) Current assets, (3) Non-current liabilities, (4) Current liabilities. Which formula correctly calculates capital?
Option C is correct: Capital = Non-current assets + Current assets − Non-current liabilities − Current liabilities
From the accounting equation: Capital = Total Assets − Total Liabilities
Total Assets = (1) + (2); Total Liabilities = (3) + (4)
Therefore: Capital = 1 + 2 − 3 − 4
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