All categories

4 years ago

What causes long sightedness?how is it corrected.

Physics

1 answer:

NNADVOKAT [17]4 years ago

4 0

Answer:

Long sight occurs when the eyeball is too short or the lens is too thin, or both. As a result, light rays from near objects are focused behind the retina because the light rays are not converged enough. The image formed on the retina is therefore out of focus.

To correct this problem, people can wear glasses with convex lenses. Light rays from near objects are converged by the convex lenses before entering the eyes, so that light can be focused on the retina to form a sharp image. Additionally, long sight can also be corrected by surgical methods such as LASIK.

You might be interested in

13.2.2 7+6 - 9(-2)+(-3)

aniked [119]

Answer:

34.27 is the answer

4 0

3 years ago

A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

bekas [8.4K]

Answer:

Explanation:

Given that,

Radius r = 15cm = 0.15m

Area of the circular loop can be determined using the formula for area of a circle

A = π r²

A = π × 0.15²

A = 0.0708 m²

Magnetic field B = 1.2T in positive z direction

B = 1.2 •k T.

If loop is remove from the field in the time interval

∆t = 2.3ms = 2.3×10^-3s

We want to find the average EMF and it is given as

ε = —∆Φ/∆t

The final flux is zero

Φf = 0

Where magnetic flux is given as

Φi = BA Cosθ

Where θ=0 since the area and the magnetic field point in the same direction.

Φi = BA Cos0

Φi = BA

Φi = 1.2 × 0.0708

Φi = 0.0848 Vs

Then, ε = —∆Φ/∆t

ε = —(Φf — Φi) / ∆t

ε = —(0-0.0848) / (2.3×10^-3)

ε = 0.0848 / (2.3×10^-3)

ε = 36.88 V

The EMF is 36.88 Volts

6 0

4 years ago

An object initially at rest experiences an acceleration of 9.8 m/s2 how much time will it take it to achieve a velocity of 58 m/

Ber [7]

Answer:

The time required by object to achieve velocity 58 m/s is 5.918 second.

Given:

acceleration = 9.8 $\frac{m}{s^{2} }$

Initial velocity = 0 m/s

final velocity = 58 m/s

To find:

Time required by object = ?

Formula used:

According to first equation of motion is given by,

v = u + at

Where, v = final velocity

u = initial velocity = 0 (Given The particle is at rest initially)

a = acceleration

t = time

Solution:

According to first equation of motion is given by,

v = u + at

Where, v = final velocity

u = initial velocity = 0 (Given The particle is at rest initially)

a = acceleration

t = time

58 = 0 + 9.8 (t)

t = $\frac{58}{9.8}$

t = 5.918 s

The time required by object to achieve velocity 58 m/s is 5.918 second.

8 0

4 years ago

A particle moves on a line away from its initial position so that after t hours it is s = 6t2 + 2t miles from its initial positi

docker41 [41]

<span>32 mph
First, let's calculate the location of the particle at t=1, and t=4
t=1
s = 6*t^2 + 2*t
s = 6*1^2 + 2*1
s = 6 + 2
s = 8 t = 4
s = 6*t^2 + 2*t s = 6*4^2 + 2*4
s = 6*16 + 8
s = 96 + 8
s = 104
So the particle moved from 8 to 104 over the time period of 1 to 4 hours. And the average velocity is simply the distance moved over the time spent. So:
avg_vel = (104-8)/(4-1) = 96/3 = 32
And since the units were miles and hours, that means that the average speed of the particle over the interval [1,4] was 32 miles/hour, or 32 mph.</span>

6 0

3 years ago

42,43 and 44 please

Mariana [72]

42) The sailboat travels east with velocity $v_e=30 mph$ , while the current moves south with speed $v_s=30 mph$ . Since the two velocities are perpendicular to each other, he resultant velocity will be given by the Pytagorean theorem:
$v= \sqrt{v_e^2+v_s^2}= \sqrt{(30)^2+(30)^2}=42 mph$
and the direction is in between the two original directions, therefore south-east. So, the correct answer is
D) 42 mph southeast

43) Since the light moves by uniform motion, we can calculate the distance corresponding to one light year by using the basic relationship between velocity, space and time. In fact, we know the velocity:
$v=300,000,000 m/s=3 \cdot 10^8 m/s$
and the time is one year, corresponding to:
$t=32,000,000 s=3.2 \cdot 10^7 s$
therefore, the distance corresponding to one light year is:
$S=vt=(3\cdot 10^8 m/s)(3.2 \cdot 10^7 s)=9.6 \cdot 10^{15}m$
Therefore, the correct answer is D.

44) For the purpose of the problem, we can assume that the light travels instantaneously from the flash to us (because the distances involved are very small), so the time between the flash and the thunder corresponds to the time it took for the sound to travel to us.
The speed of sound is
$v=1100 ft/s=750 mph=335 m/s$
And since the time between the flash and the thunder is t=3 s, the distance is
$S=vt=(335 m/s)(3 s)=1005 m=0.62 miles$
Therefore, the correct answer is A) 3/5 mile.

6 0

3 years ago