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zloy xaker [14]
3 years ago
12

The eyes of many older people have lost the ability to accommodate, and so an older person’s near point may be more than 25 cm f

rom the eye. Does the older person’s eye have a larger or smaller refractive power than a normal eye, when focused on an object at the near point? Explain.
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer:

Smaller refractive power

Explanation:

The refractive power of an eye is the extent to which it can converge or diverge the light rays.

Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.

A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.

Thus the near point also increases and the refractive power becomes smaller.

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Plants exhibit phototropism because _____. long roots help plants absorb more water and nutrients they need the Sun's energy in
vlada-n [284]

they need the Sun's energy in order to produce food <span>
There are several starting ingredients that are required for starting the process of photosynthesis. The most important are sunlight, water, carbon dioxide and certain minerals. Sunlight helps in building up the optimum temperature that is required to start the process of photosynthesis. Water is an important raw material that helps to absorb the carbon dioxide from the air. It also directly influences the opening and closing of the stomata. Carbon dioxide on the other hand is one of the most important raw material that helps to make glucose via the process of photosynthesis.</span>

6 0
3 years ago
The two vectors and in fig. 3-28 have equal magnitudes of 10.0 m and the angles are 30° and 105°. find the (a) x and (b) y compo
mylen [45]

You can just use basic trigonometry to solve for the x & y components.

<span>vector a = 10cos(30) i + 10sin(30) j = <5sqrt(3), 5></span>

vector b is only slightly harder because the angle is relative to vector a, and not the positive x-axis. Anyway, this just makes vector b with an angle of 135deg to the positive x-axis.

<span>vector b = 10cos(135) i + 10sin(135) j = <-5sqrt(2), 5sqrt(2)></span>

So now we can do the questions:

r = a + b

r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>

(a) 5sqrt(3)-5sqrt(2)

(b) 5+5sqrt(2)

(c)

|r| = sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )

= 12.175

(d)

θ = tan-1 ( (5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )

θ = 82.5deg

<span> </span>

6 0
3 years ago
The Demon Drop ride at the amusement park falls freely for 1.8 s after
timofeeve [1]

Answer:

Explanation:

a = -g = -9.80 m/s squared

d o  = 0

v o =  0

t = 1.8s

<h3>unkown:</h3>

d = ?

v = ?

8 0
3 years ago
Read 2 more answers
What causes a compass needle to point to geographic north?
morpeh [17]
I believe it’s “C”
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8 0
2 years ago
A gang of robbers is escaping across city roofs at night. They come to the edge of one building and need to drop down to their g
REY [17]

Answer:

a) They will hit the ground with a speed of 19.6 m/s.

b) They are at a height of 20 m.

c) It is not a safe jump.

Explanation:

Hi there!

a) The equations of height and velocity in function of time of a free falling body are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the object at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering downward as negative direction).

v = velocity of the object at time t.

Using the equation of velocity, let's find the velocity at which they will hit the ground. The pebble is dropped (initial velocity = 0) and it takes 2 s to reach the ground:

v = v0 + g · t     (v0 = 0)

v = g · t

v = -9.8 m/s² · 2.0 s

v = -19.6 m/s

They will hit the ground with a speed of 19.6 m/s.

b)Now, we have to use the equation of height:

h = h0 + v0 · t + 1/2 · g · t²

If we place the origin of the frame of reference on the ground, we have to find the initial height (h0) knowing that at t = 2.0 s, h = 0 m

0 m = h0 - 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 1/2 · 9.8 m/s² · (2.0 s)²

h0 = 20 m

They are at a height of 20 m.

c)According to a NASA paper (Issues on Human Acceleration Tolerance After Long-Duration Space Flights, figure 10), if you fall with a vertical velocity greater than 17 m/s it is unlikely that you will survive. So, it is not a safe jump.  

3 0
3 years ago
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