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2 years ago

Which option do you think could cause near sightedness?

Physics

2 answers:

Softa [21]2 years ago

7 0

Answer:A

Explanation:because it doesn’t refract light so you can’t see close

Travka [436]2 years ago

6 0

Answer:

A) the eye lens doesn't refract light

Explanation: because the eye doesn’t refract, or bend, light properly to focus on the retina. Makes eyesight appear blurry and distorted.

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Algunas fabricas de balones de fútbol ubicadas en la costa inflan los balones que van a ser vendiéndose las ciudades como pasto,

otez555 [7]

Answer:

balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure

Explanation:

Este ejercicio es referente a la mecánica de fluidos, usemos la expresión para la presión

P = ρ g h

En es el caso del balón usemos la presión en la pared extrema, llamemos P la presión por el gas en el interior y P_ext la presión atmosférica del lugar

cuando se llena el valor en una ciudad de baja altura la presión atmosférica es mas alta

P_int1 < P_ext1

por lo cual la pared del balón no se mantiene rígida.

Cuando el balón es trasladado a una ciudad con mayor altura sobre el nivel del mar la presión exterior disminuye

P_ext2 = ρ g h₂ < P_ext1

en promedio la presión disminuye con la altura en 0,029 atm cada 250 m

por lo tanto como la cantidad de gas es constante en el interior la presión interior del globo es constante y en esta ciudades se hace igual o un poco mayor que la presión atmosférica, en consecuencia la pared del globo esta rígida

P_int2 >P_ext2

Traslate

This exercise is related to fluid mechanics, let's use the expression for pressure

P = ρ g h

In the case of the balloon, let's use the pressure on the extreme wall, let's call P the pressure for the gas inside and P_ext the atmospheric pressure of the place

when the value is filled in a low-lying city the atmospheric pressure is higher

P_int1 <P_ext1

therefore the wall of the ball does not remain rigid.

When the ball is transferred to a city with higher altitude above sea level, the external pressure decreases

P_ext2 = ρ g h <P_ext1

on average the pressure decreases with height by 0.029 atm every 250 m

therefore as balloon is rigid the amount of gas is constant inside the interior pressure of the balloon is constant and in these cities it becomes equal to or slightly higher than atmospheric pressure, therefore the wall of the

Pint 2> Pe

8 0

3 years ago

When a liquid is introduced into the air space between the lens and the plate in a Newton's-rings apparatus, the diameter of the

Leviafan [203]

Answer:

$n_l=1.58$

Explanation:

Newton's ring equation relates the initial and final diameter of the ring, to the index of refraction of the air and the index of refraction of the liquid between the lens and the plates, as follows:

$n_ld_f^2=n_ad_i^2$

Solving for :

$n_l=n_a\frac{d_i^2}{d_f^2}\\$

Recall that the refractive index of the air is 1. So, replacing the given diameters:

$n_l=\frac{(1.52cm)^2}{(1.21cm)^2}\\\\n_l=1.58$

3 0

4 years ago

A block of mass 10 kg and measuring 250 mm on each edge is pulled up an inclined surface on which there is a film of SAE 10W-30

Ivan

Answer:

a) 2.53 * 10^-2 m/s

b) -4.78 * 10^-2 m/s

c) 1.21 * 10^-1 m/s

Explanation:

Given data :

Mass of block = 10 kg

Measuring 250mm on each side

a) calculate the speed when a force of 75N is applied to pull block upwards

F = f + W sin∅ ( equation for applying the force of equilibrium condition in the x axis ) ----- ( 1 )

f ( friction force )= ( 16400v * 6.25 *10^-2) = 1025 v

F ( force applied ) = 75

W ( weight of block ) = 10 * 9.81 = 98.1 N

∅ = 30°

input values into equation 1

V = $\frac{75- (98.1*sin30^{0}) }{1025}$ = 2.53 * 10^-2 m/s

b) Speed when no force is applied on the block

F = f + W sin∅

F = 0

f = 1025 V

W = 98.1 N

∅ = 30°

hence V = $\frac{0 - (98.1*sin30^{0}) }{1025}$ = - 4.78 * 10^-2 m/s

c) when a force is applied to push block down the incline

F = f + W sin∅ ----- ( 3 )

F = 75 N

f = 1025 V

W = 98.1 N

∅ = 30°

input values into equation 3 considering the fact that the weight of the block is acting in the opposite direction

75 = 1025 V - 98.1 ( sin 30° )

V = $\frac{75+( 98.1*sin30^{0}) }{1025}$ = 1.21 * 10^-1 m/s

5 0

3 years ago

In a rigid container, when the speed of the gas molecules decreases, the pressure of the gas also decreases.

Paraphin [41]

True please mark brainly

8 0

4 years ago

The efficiency of a carnot cycle is 1/6. If on reducing the temperature of the sink 75 degree Celsius, the efficiency becomes 1/

svp [43]

Answer:

375 and 450

Explanation:

The computation of the initial and the final temperature is shown below:

In condition 1:

The efficiency of a Carnot cycle is $\frac{1}{6}$

So, the equation is

$\frac{1}{6} = 1 - \frac{T_2}{T_1}$

For condition 2:

Now if the temperature is reduced by 75 degrees So, the efficiency is $\frac{1}{3}$

Therefore the next equation is

$\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}$

Now solve both the equations

solve equations (1) and (2)

$2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)$

T_2 + 450 = 75

T_2 = 375

Now put the T_2 value in any of the above equation

i.e

T_1 = T_2 + 75

T_1 = 375 + 75

= 450

7 0

4 years ago