Answer:
The work done on the hose by the time the hose reaches its relaxed length is 776.16 Joules
Explanation:
The given spring constant of the of the spring, k = 88.0 N/m
The length by which the hose is stretched, x = 4.20 m
For the hose that obeys Hooke's law, and the principle of conservation of energy, the work done by the force from the hose is equal to the potential energy given to the hose
The elastic potential energy, P.E., of a compressed spring is given as follows;
P.E. = 1/2·k·x²
∴ The potential energy given to hose, P.E. = 1/2 × 88.0 N/m × (4.20 m)²
1/2 × 88.0 N/m × (4.20 m)² = 776.16 J
The work done on the hose = The potential energy given to hose, P.E. = 776.16 J
Answer: Answer is D
I took the test little while back.
Answer: Carbon tetrachloride Or Tetrachloromethane
Explanation: Carbon tetrachloride is an important nonpolar covalent compound. You determine its name based on the atoms present in the compound.
<span>won
adjective
Verb phrases are verbs that may function as a predicate, adjective, or adverb. </span>
(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.
<span>(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. </span>
<span>(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. </span>
<span>(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.</span>
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
