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2 years ago

Determine the focal length of a plano-concave lens (refractive index n =1.5) with 24 cm radius of curvature on its curve surface

Physics

1 answer:

tatyana61 [14]2 years ago

3 0

Answer:

Option 3: -48 cm

Explanation:

We are given:

refractive index; n = 1.5

radius of curvature; r2 = 24 cm

Formula for the focal length is given as;

1/f = (n - 1) × [(1/r1) - (1/r2)]

As r1 tends to infinity, 1/r1 = 0

Thus,we now have;

1/f = (n - 1) × (-1/r2)

Plugging in the relevant values;

1/f = (1.5 - 1) × (-1/24)

1/f = -0.02083333333

f = -1/0.02083333333

f = -48 cm

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A freight train rolls along a track with considerable momentum. If it were to roll at the same speed but had twice as much mass,

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Answer:

The momentum would be doubled

Explanation:

The magnitude of the momentum of the freight train is given by:

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v is its speed

In this problem, we have that the speed of the train is unchanged, while the mass of the train is doubled:

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3 years ago

A diver leaves the end of a 4.0 m high diving board and strikes the water 1.3s later, 3.0m beyond the end of the board. Consider

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Answer:

4.0 m/s

Explanation:

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$d=v_x t$

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d = 3.0 m is the horizontal distance covered

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Now let's consider the vertical motion: this is an accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. The vertical position at time t is given by

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where

h = 4.0 m is the initial height

vy is the initial vertical velocity

We know that at t = 1.3 s, the vertical position is zero: y = 0. Substituting these numbers, we can find vy

$0=h+v_y t - \frac{1}{2}gt^2\\v_y = \frac{0.5gt^2-h}{t}=\frac{0.5(9.8 m/s^2)(1.3 s)^2-4.0 m}{1.3 s}=3.3 m/s$

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3 years ago

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2 years ago

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The ratio of the focal length of a lens to it's diameter is the

user100 [1]

Answer:

The f-ratio describes the relationship between the lens diameter and the focal length and is calculated by dividing the focal length by the diameter of the lens. For example, if a lens were to have a focal length of 50mm and a diameter of 10mm, then the f-ratio would be 50mm/10mm=5 or otherwise referred to as f5.

Explanation:

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2 years ago

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We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o

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The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T ...(i)

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Resultant force m₁a = m₁g sin α+T

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Solving both the equations by adding them,

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18a=11gsin 65°+7g sin 15°=115.45

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2 years ago