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Natali5045456 [20]
3 years ago
12

John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds

it difficult to express his thoughts and cannot seem to find the right words to say while speaking. However, he can speak freely with proper syntax. In this scenario, John is most likely suffering from _______.
Physics
1 answer:
katovenus [111]3 years ago
6 0

Answer: ​Wernicke's aphasia

Explanation:

John recently suffered a blow to his head. Since then, he finds it difficult to comprehend what others say to him. He also finds it difficult to express his thoughts and cannot seem to find the right words to say while speaking. However, he can speak freely with proper syntax. In this scenario, John is most likely suffering from Wernicke's aphasia.

Wernicke's aphasia occurs when the leftward side of the middle of the brain is damaged or has been altered. An individual who suffers from Wernicke's aphasia will have difficulty in speaking in meaningful and coherent sentences or may have difficulty in understanding the speech of others.

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Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

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