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lesya [120]
2 years ago
11

Two convex thin lenses with focal lengths 10.0 cm and 20.0 cm are aligned on a common axis, running left to right, the 10-cm len

s being on the left. A distance of 20.0 cm separates the lenses. An object is located at a distance of 15.0 cm to the left of the 10-cm lens. Where will the final image appear as measured from the 20-cm lens? (a) -13.3 cm (b) -6.67 cm (c) +6.67 cm (d) +13.3 cm (e) +8.1 cm
Physics
1 answer:
love history [14]2 years ago
3 0

Answer:

(c) +6.67

Explanation:

f1 = 10 cm

f2 = 20 cm

u = Object distance = 15 cm

Distance between lenses = 20 cm

For first lens image distance

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{10}-\frac{1}{15}\\\Rightarrow \frac{1}{v}=\frac{1}{30}\\\Rightarrow v=30\ cm

Distance from second lens is 10 cm to the right

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{20}-\frac{1}{-10}\\\Rightarrow \frac{1}{v}=\frac{3}{20}\\\Rightarrow v=6.67\ cm

The final image will appear as +6.67 cm

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Which of the following statements is false?
agasfer [191]

A & B

Observe the path of the light ray as it passes through the lenses as shown in the attached images. Concave lenses diverge light rays while the convex lens converges the light rays.

Explanation:

Real images are formed where the rays converge, a property of images by convex lenses. Convex lenses can be used to magnify objects. If the image occurs before the focal point of the lens then the image will be upright but smaller. The images inverts and gets bigger past the focal point.

Virtual images are property of concave lenses. These images appear closer but smaller than the real object.

Learn More:

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6 0
2 years ago
Read 2 more answers
The cockroach Periplaneta americana can detect a static electric field of magnitude 8.50 kN/C using their long antennae. If the
otez555 [7]

Answer:

0.647 nC

Explanation:

The force experienced by a charge due to the presence of an electric field is given by

F=qE

where

q is the charge

E is the magnitude of the electric field

In this problem, each antenna is modelled as it was a single point charge, experiencing a force of

F=5.50\mu N = 5.50\cdot 10^{-6} N

Therefore, if the electric field magnitude is

E=8.50 kN/C = 8500 N/C

Then the charge on each antenna would be

q=\frac{F}{E}=\frac{5.50\cdot 10^{-6} N}{8500 N/C}=6.47\cdot 10^{-10} C = 0.647 nC

8 0
2 years ago
A reflecting telescope is built with a 20-cm-diameter mirror having a 1.00 m focal length. it is used with a 10× eyepiece.
kondaur [170]
I really wish I could be helping you. I don't know.
3 0
2 years ago
Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

6 0
2 years ago
The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is
Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

            W = ΔEm

Where the work is

             W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

           Em₀ = K = ½ m v²

Final. When the system is at rest

            Em_{f} = K_{e1}b +K_{e2} = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

            ∑ F = F - fr

Axis y

           N- W = 0

           N = W

The friction force has the equation

          fr = μ N

          fr = μ W

  The job

         W = (F – μ W) x

We substitute in the equation

            (F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

           ½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values ​​and solve

           ½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

         x² 30 + 14.4 x - 1,125 = 0

        x² + 0.48 x - 0.0375 = 0

           

We solve the second degree equation

        x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

        x = [-0.48 ± 0.617] / 2

        x₁ = 0.0685 m

        x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0
2 years ago
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