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3 years ago

Defects of vision and their correction drawing No Spams ❌❌

Physics

2 answers:

Vinvika [58]3 years ago

6 0

Answer:

There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness (ii) Hypermetropia or far – sightedness (iii) Presbyopia. Myopia is also known as near-sightedness. A person with myopia can see nearby objects clearly but cannot see distant objects distinctly.

Explanation:

Ahat [919]3 years ago

6 0

Answer:

Vision Correction

You probably know people who need eyeglasses or contact lenses to see clearly. Maybe you need them yourself. Lenses are used to correct vision problems. Two of the most common vision problems are myopia and hyperopia.

Explanation:

hope this helps

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Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars. (T/F)

-BARSIC- [3]

Answer:

Astronomers have no theoretical explanation for the ""hot Jupiters"" observed orbiting some other stars.

False

Explanation:

The “hot Jupiters” joint word startes to be used to be able to describe planets like 51 Pegasi b, a planet with a 10-day-or-less orbit and a mass 25% or greater than Jupitere, circling a sun-like star planet in 1995, which was found by astronomers Michel Mayor and Didier Queloz, who were awarded the 2019 Nobel Prize for Physics along with the cosmologist James Peebles for their “contributions to our understanding of the evolution of the universe and Earth’s place in the cosmos.”

Now we know a total of 4,000-plus exoplanets, but only a few more than 400 meet the definition of the enigmatic hot Jupiters which, tell us a lot about how planetary systems form, and what kinds of conditions cause extreme results.

In a 2018 paper in the Annual Review of Astronomy and Astrophysics, astronomers Rebekah Dawson of the Pennsylvania State University and John Asher Johnson of Harvard University reviewed on how hot Jupiters might have formed, and would be the meaning for the rest of the planets in the galaxy.

4 0

3 years ago

Water enters a student's house 10.0 m above the ground through a pipe with a cross section area of 1.00 x 10-4m2 at ground. Insi

dezoksy [38]

Answer:

(a). V₁ = 10m/s (velocity inside the house), V₂ = 5m/s (velocity at ground level)

(b). P₂ = 236500 Pa

Explanation:

This is quite straight-forward so let us begin by defining the terms given.

Given that;

The cross-section area inside the student's house A₁ = 0.50 0.50 x 10-4m2.

Let us make the velocity of water inside the house be V₁

such that the Volume of water entering the per second is = A₁V₁

Therefore, in 90sec:

45 L = 90 A₁V₁

V₁ = 45 * 10⁻³m³ / 90*0.5*10⁻⁴

V₁ = 10m/s (velocity of water inside the house)

From the continuity equation we have that;

A₁V₁ = A₂V₂

0.5*10⁻⁴ * 10 = 1*10⁻⁴ V₂

V₂ = 5m/s (velocity at ground level)

(b). We are told to calculate the water pressure in the pipeline at the ground level.

Using Bernoulli's equation;

P₁ + pgh₁ + 1/2PV₁² (inside) = P₂ + pgh₂ + 1/2PV₂² (ground level)

1.01*10⁵ + 1000*9.8*10 + 1/2*1000*(10)² = P₂ + 0 + 1/2*1000*(5)²

P₂ (pressure) = 1.01*10⁵Pa

Therefore we have;

101000 + 98000 + 50000 = P₂ + 12500

P₂ = 236500 Pa

cheers I hope this helped !!

3 0

3 years ago

which of the folloing statements about ionization energy is true? A elements toward the bottom of a group periodic table general

beks73 [17]

<span>
accept the flow of electrons.resist the flow of electrons.accept the flow of protons.resist the flow of protons.It is one of these </span>

6 0

3 years ago

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, in

olya-2409 [2.1K]

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=\frac{I}{\Delta t}=\frac{328.6 kg m/s}{0.812 s}=404.7 N$

7 0

3 years ago

Calculate the distance travel if it accelerates from 0 to 27.8 meters per second in 2.5 seconds

luda_lava [24]

The distance travel is 69.5 meters.

<u>Explanation:</u>

Given datas are as follows

Speed = 27.8 meters / second

Time = 2.5 seconds

The formula to calculate the speed using distance and time is

Speed = Distance ÷ Time (units)

Then Distance = Speed × Time (units)

Distance = (27.8 × 2.5) meters

Distance = 69.50 meters

Therefore the distance travelled is 69.50 meters.

8 0

3 years ago

Defects of vision and their correction drawing ​No Spams ❌❌

Defects of vision and their correction drawing No Spams ❌❌