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UkoKoshka [18]
3 years ago
13

If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t

he person would need to see an object at infinity clearly?
Physics
1 answer:
Anastaziya [24]3 years ago
8 0

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

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Soap bubble interference Light of 690-nm wavelength interferes constructively when reflected from a soap bubble having refractiv
Nikitich [7]

Answer:

The two possible thicknesses of the soap bubble is 129 nm and 389 nm.

Explanation:

Given that,

Wavelength = 690 nm

Refractive index = 1.33

We need to calculate the two possible thicknesses of the soap bubble

Using formula of thickness

t=\dfrac{(m+\dfrac{1}{2})\lambda}{2n}

For m = 0,

Put the value into the formula

t=\dfrac{\dfrac{690\times10^{-9}}{2}}{2\times1.33}

t=129\times10^{-9}\ m

t=129\ nm

For m=1,

Put the value into the formula

t=\dfrac{\dfrac{3\times690\times10^{-9}}{2}}{2\times1.33}

t=389\times10^{-9}\ m

t=389\ nm

Hence, The two possible thicknesses of the soap bubble is 129 nm and 389 nm.

5 0
3 years ago
How many significant figures is in 15,600
amid [387]
3 significant figures.
5 0
3 years ago
Which of the following biomes might be found at a latitude of 33° South?
cupoosta [38]
The answer is going be desert. 
5 0
3 years ago
Now we’ll use the component method to add two vectors. We will use this technique extensively when we begin to consider how forc
dimaraw [331]

Answer:

The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Explanation:

Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.

Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.

So, the vector sum R = A + B

The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm

The y-component of the vector sum is R" = A"+ B" = 25 cm + 32.89 cm = 57.89 cm

So, the magnitude of R = √(R'² + R"²)

= √((31.33 cm)² + (57.89 cm)²)

= √(981.5689 cm² + 3,351.2521 cm²)

= √(4,332.821 cm²)

= 65.82 cm

≅ 65.8 cm

The direction of R is Ф = tan⁻¹(R"/R')

= tan⁻¹(57.89 cm/31.33 cm)

= tan⁻¹(1.84775)

= 61.58°

≅ 61.6°

So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

4 0
3 years ago
At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f
jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp =  change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s,  0.65 kg m/s)

The magnitude of this vector is calculated as follows:

|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0
3 years ago
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