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Debora [2.8K]
3 years ago
15

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

Physics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}

Where;

v = The image distance from the lens

We get;

v = \dfrac{u \times f}{u - f}

Therefore;

v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

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Blababa [14]

Answer:

a) time taken to catch up with speeding car is 12.25 secs

b) the police car will travel 273.8 m to catch up with the speeding car

Explanation:

Given that;

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acceleration of police car = 10 mi/hr = 4.47 m/s²

V_{f}  = 70 mi/hr = 31.29 m/s

Now time taken to reach maximum speed is t₁

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V_{f} =  V_{i} + at₁

we substitute

31.29 = 0 + 4.47t₁

t₁ = 31.29 / 4.47

t₁  = 7 sec

now

d₁ = 0 + 1/2 × at₁²

d₁ = 0 + 1/2 × 0 + 4.47×(7)²

d₁ = 109.5 m

so distance travelled by the speeding car in time t₁  will be

d_{c} = V_{c} × t₁

we substitute

d_{c} = 22.352 × 7

d_{c}  = 156.46 m

now distance between polive car and speeding car

Δd =  d_{c} - d₁

Δd = 156.46 - 109.5

Δd = 46.96 m

time taken to cover Δd will be

t₂ = Δd / ( V_{f} - V_{c} )

t₂ = 46.96 / ( 31.29 - 22.352 )

t₂ = 46.96 / 8.938

t₂ = 5.25 sec

distance travelled by the police in time t₂ will be

d₂ = V_{f} × t₂

d₂ = 31.29 × 5.25

d₂ = 164.3 m

a) How long will it take before the officer catches up to the speeding car;

time taken to catch up with speeding car;

t = t₁ + t₂

t = 7 + 5.25

t = 12.25 secs

Therefore, time taken to catch up with speeding car is 12.25 secs

b)  how far will it have travelled in order to do so;

distance = d₁ + d₂

distance = 109.5 + 164.3

distance = 273.8 m

Therefore, the police car will travel 273.8 m to catch up with the speeding car

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An electromagnetic wave has a frecuency of 4.0 x 10 ¹⁸Hz and the speed of light is 3.0 x 10⁸ m/s. So:

λ = (3.0 x 10⁸ m/s)/(4.0 x 10¹⁸ Hz)

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