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ArbitrLikvidat [17]
3 years ago
12

Sometimes a person cannot clearly see objects close up or far away. To correct this type of vision, bifocals are often used. The

top half of the lens is used to view distant objects and the bottom half of the lens is used to view objects close to the eye. A person can clearly see objects only if they are located between 34 cm and 180 cm away from his eyes. Bifocal lenses are used to correct his vision. What power lens (in diopters) should be used in the top half of the lens to allow him to clearly see distant objects
Physics
1 answer:
Alexxx [7]3 years ago
3 0

Answer:

The power of top half of the lens is 0.55 Diopters.

Explanation:

Since, the person can see an object at a distance between 34 cm and 180 cm away from his eyes. Therefore, 180 cm must be the focal length of the upper part of lens, as the top half of the lens is used to see the distant objects.

The general formula for power of a lens is:

Power = 1/Focal Length in meters

Therefore, for the top half of the lens:

Power = 1/1.8 m

<u>Power = 0.55 Diopters</u>

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Do you think seismographs predict earthquakes or measure earthquakes, explain your answer?
Yuki888 [10]

Answer:

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Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

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4 0
4 years ago
A package of aluminum foil contains 50. Ft2 of foil, which weighs approximately 7.0 oz . Aluminum has a density of 2.70 g/cm3. W
kumpel [21]

We know density = Mass / Volume

So Volume = Mass/Density

Volume =  Area * Thickness

Area = 50 ft^2 = 4.645152 m^2\\ \\Mass = 7 oz = 0.198447 kg\\ \\Density = 2.70 g/cm^3 = 2700 kg/m^3\\ \\Substituting \\ \\4.645152 *t = \frac{0.198447}{2700} \\ \\t = 1.58*10^{-5}m = 1.58*10^{-2}mm

So the approximate thickness of the foil in millimeters = 1.58*10^{-2}mm

6 0
4 years ago
A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the bas
trasher [3.6K]

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

t= \sqrt{\frac{2*37.5}{9.8} }

t = 2.77 s

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Consider a piston filled with 3 mols of an ideal gas, kept at a constant temperature 290 K. We slowly compress the gas starting
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Answer: -5013.65\ J

Explanation:

Given

No of moles n=3

Temperature T=290\ K

Initial volume V_1=2\ m^3

Final volume V_2=1\ m^3

Work done in constant temperature process is

W=nRT\ln \left(\dfrac{V_2}{V_1}\right)

Insert the values

\Rightarrow W=3\times 8.314\times 290\ln \left (\dfrac{1}{2}\right)\\\\\Rightarrow W=-870\times 8.314\times \ln (2)\\\Rightarrow W=-5013.65\ J

5 0
3 years ago
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