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levacccp [35]
3 years ago
5

How does the distance from an eye chart affect the number of letters that are recognized on a line

Physics
2 answers:
luda_lava [24]3 years ago
8 0

Explanation:

Distance from the eye chart are used to find defect in the in the eye sights of a patient. If one cannot see far off things, that means he is near sighted. So if he goes away from the eye chart and try to see the letters in the chart, he will see blur image of the letters.

           And if one is far sighted he cannot see the near by things. If one comes closer to the chart and tries to read the letters then he will not be able to read those letters because he is far sighted.

Thus, in this way distance from the eye chart affect to recognize the letters present in the eye chart.

Nat2105 [25]3 years ago
3 0
Well, if you are nearsighted which means you can't see far, the father you are the more blurry it will be, resulting to you saying the wrong letter. This then shows the doctor that you can't see far away, nearsighted.
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Why do we feel the force of gravity from the Earth but we don’t feel the force of gravity from the moon?
Zigmanuir [339]

Answer:

The gravitational force on the moon is less than on Earth because the strength of gravity is determined by an object's mass. The bigger the object, the bigger the gravitational force. Gravity is pretty much everywhere. We just feel it in different ways depending on our state of motion.

Explanation:

Hope this helped!!

8 0
2 years ago
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A star is born when gas and dust from a nebula become so dense and hot that nuclear fusion starts. Which of the following forces
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8 0
3 years ago
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Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in p
irina1246 [14]

Answer:

1 ohm

Explanation:

since there are two identical resistors, one resistor will be

R = \frac{4}{2} =2ohm [ proven as in series R_{e} = 2 + 2 = 4ohm ]

to calculate the equivalent resistance when in parallel:

\frac{1}{R_{e} }  = \frac{1}{R_{1} } + \frac{1}{R_{2}}

so,

\frac{1}{R_{e} } = \frac{1}{2} + \frac{1}{2}

R_{e} = 1ohm

4 0
2 years ago
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron
slega [8]

Answer:

a)   v_average = 11 m / s, b)  t = 0.0627 s

, c)    F = 7.37 10⁵ N

, d)   F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

The average speed is

           v_average = (v + v₀) / 2

           v_average = (22 + 0) / 2

           v_average = 11 m / s

b) The average time

          v = v₀ - a t

          t = v₀ / a

          t = 22 / 350.72

          t = 0.0627 s

c) The force can be found with Newton's second law

             F = m a

             F = 2100 350.72

             F = 7.37 10⁵ N

.d) the ratio of this force to weight

             F / W = 7.37 10⁵ / (2100 9.8)

             F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

5 0
3 years ago
A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

where v is the speed, T is the tension and \mu is the mass per unit length.

Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

4 0
3 years ago
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