13. Mohan has a box of coloured pens. He takes a pen at random from the box. The probability that he takes a red pen is 0.4. If the box contains total 50 pens of blue, green and red colours, and there are 15 blue pens and some green pens, then answer the following :(1) The no of red pen in the box are:(a) 15(b) 20(c) 25(ii) The probability of taking a green pen is:(a) 0.6(b) 0.4(c) 0.5(iv) The probability of taking a blue or green pen is(a) 0.3(b) 0.4(c) 0.5(d) 0.6 ​

Answer:

solutions are of 2 types

Answer:

Cleansing action of soaps: ... When soap i dissolved in water, its hydrophobic ends attach themselves to the dirt and remove it from the cloth. Then, the molecules of soap arrange themselves in micelle formation and trap the dirt at the centre of the cluster. These micelles remain suspended in the water.

Explanation:

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A linguistic expression shows semantic ambiguity when it can have multiple senses, at least when uttered out of context. Lexical ambiguity is the subtype of semantic ambiguity which occurs at the level of words or morphemes. When a lexical ambiguity results from a single word having two senses, it is called polysemy (e.g. the "foot" of a person versus the "foot" of a pot). When it instead results from two separate words which happen to be pronounced the same way, it is called homonymy. In practice, these types of lexical ambiguity can be difficult to distinguish.

Semantic ambiguity can result from a sentence allowing multiple ways of semantically composing its constituent expressions. Scope ambiguity and de re/de dicto ambiguities are two notable examples of this kind of ambiguity. In some frameworks, such ambiguities are the semantic reflexes of syntactic ambiguities, though in other approaches they are not.

Answer:

Refer above attachment.

• Two charges of 2uC and 8 uC are separated by 4 cm.

• Electrostatic force between the charges?

• Force when distance is halved and a dielectric constant 2 is placed?

• Change in force ?

Initial force is :

[tex]F = \dfrac{k(q1 \times q2)}{ {r}^{2} } [/tex]

[tex] \implies F = \dfrac{(9 \times {10}^{9}) (2 \times 8 \times {10}^{ - 12} )}{ {(0.04)}^{2} } [/tex]

[tex] \implies F = \dfrac{(9 \times {10}^{9}) (16 \times {10}^{ - 12} )}{ 16 \times {10}^{-4} } [/tex]

[tex] \implies F = 9 \times {10}^{9} \times {10}^{ - 8} [/tex]

[tex] \implies F = 90 N[/tex]

So, force is 90 N.

In 2nd case :

[tex]F2 = \dfrac{k(q1 \times q2)}{K \times {r}^{2} } [/tex]

[tex] \implies F2 = \dfrac{(9 \times {10}^{9}) (2 \times 8 \times {10}^{ - 12} )}{ 2 \times {(0.02)}^{2} } [/tex]

[tex] \implies F2 = \dfrac{(9 \times {10}^{9}) (16\times {10}^{ - 12} )}{ 8 \times {10}^{-4} } [/tex]

[tex] \implies F2 = 9 \times {10}^{9} \times 2\times {10}^{ - 8} [/tex]

[tex] \implies F2 = 180 N[/tex]

So, change in force is :

[tex]\Delta F = (180- 90) [/tex]

[tex] \implies \Delta F = 90 N[/tex]

Hope It Helps.