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algol13
3 years ago
9

Sonia's grandfather was not able to see clearly after returning from his walk on a bright sunny day. He was about to hit a chair

when sonia held him and guided him to the nearby sofa. What could be the reason for sonia's grandfather not being able to see clearly on entering the house
Physics
1 answer:
Vinil7 [7]3 years ago
4 0

Answer:

Sonia's grandfather was not able to see clearly after returning from his walk on a bright sunny day. He was about to hit a chair when Sonia held him and guided him to the nearby sofa, in sunny environment grandfather's eye is set according to high brightness, his pupils become small to lower the number of light rays entering his eye, when he returned from sunny environment to his house which was having low brightness so the pupils should enlarge to absorb more light to see clearly but due to old age his Ciliary muscle of the eye would have worn out and due to this poor coordination the image was not clear.

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So far in your life, you may have assumed that as you are sitting in your chair right now, you are not accelerating. However, th
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Answer:

a) a=33.73mm/s^{2}

b) mg>N

c) \%_{change}=0.343\%

d) a=24.07mm/s^{2}

Explanation:

In order to solve part a) of the problem, we can start by drawing a free body diagram of the presented situation. (see attached picture).

In this case, we know the centripetal acceleration is given by the following formula:

a_{c}=\omega ^{2}r

where:

\omega=\frac{2\pi}{T}

we know the period of rotation of the earth is about 24 hours, so:

T=24hr*\frac{3600s}{1hr}=86400s

so we can now find the angular speed:

\omega=\frac{2\pi}{86400s}

\omega=72.72x10^{-6} rad/s^{2}

So the centripetal acceleration will be:

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which yields:

a_{c}=33.73mm/s^{2}

b)

In order to answer part b, we must draw a free body diagram of us sitting on a chair. (See attached picture.)

So we can do a sum of forces in equilibrium:

\sum F=0

so we get that:

N-mg+ma_{c} = 0

and solve for the normal force:

N=mg-ma_{c}

In this case, we can clearly see that:

mg>mg-ma_{c}

therefore mg>N

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c)

So let's calculate our weight and normal force:

Let's say we weight a total of 60kg, so:

mg=(60kg)(9.81m/s^{2})=588.6N

and let's calculate the normal force:

N=m(g-a_{c})

N=(60kg)(9.81m/s^{2}-33.73x10^{-3}m/s^{2})

N=586.58N

so now we can calculate the percentage change:

\%_{change} = \frac{mg-N}{mg}x100\%

so we get:

\%_{change} = \frac{588.6N-586.58N}{588.6N} x 100\%

\%_{change}=0.343\%

which is a really small change.

d) In order to find this acceleration, we need to start by calculating the radius of rotation at that point of earth. (See attached picture).

There, we can see that the radius can be found by using the cos function:

cos \theta = \frac{AS}{h}

In this case:

cos \theta = \frac{r}{R_{E}}

so we can solve for r, so we get:

r= R_{E}cos \theta

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r = (6371x10^{3}m)cos (44.4^{o})

which yields:

r=4,551.91 km

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Answer:

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