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3 years ago

I’m just completely confused

Physics

1 answer:

Neporo4naja [7]3 years ago

6 0

4. It is substances

I don't know the 3rd one

Hope it helped you

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Locate the complete verbal phrase and identify its type.

Ksju [112]

won
adjective

Verb phrases are verbs that may function as a predicate, adjective, or adverb. 

(a) "That he said" is an adjective modifying "word". However, this contains the s ubject"he" and the verb "said". It is a clause and NOT a phrase. Phrases can only have either a verb or a noun.
(b) There's only one verb "was" but it does not come with a complement, object, modifier, or other verb. Hence, it's NOT a verb phrase. 
(c) "Shall be" consists of the modal shall and the be-verb be. This is a perfect example of a verb phrase that functions as a VERB PHRASE. 
(d) "Roared" and "charged" are two verbs referring to different subjects. They do not come with a complement, object, modifier, or another verb. Hence, they're NOT a verb phrase. "As the bull charged" is a clause and not a phrase.

7 0

3 years ago

Read 2 more answers

An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto

Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

$R\propto\dfrac{1}{D}$

The power is inversely proportional to the resistance.

$P\propto\dfrac{1}{R}$

$P\propto D^2$

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

$\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}$

Put the value into the formula

$\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}$

$\dfrac{D_{i}}{D_{f}}=0.95$

$D_{i}=0.95 D_{f}$

Hence, The factor of the diameter is 0.95.

7 0

3 years ago

Read 2 more answers

After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?

faust18 [17]

Answer: Option (b) is the correct answer.

Explanation:

Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.

Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.

Thus, we can conclude that negative charge spread evenly on both ends.