Add a picture of leaves.

A screen is placed a distance dd to the right of an object. A converging lens with focal length ff is placed between the object

Elina [12.6K]

Answer:

2f

Explanation:

The formula for the object - image relationship of thin lens is given as;

1/s + 1/s' = 1/f

Where;

s is object distance from lens

s' is the image distance from the lens

f is the focal length of the lens

Total distance of the object and image from the lens is given as;

d = s + s'

We earlier said that; 1/s + 1/s' = 1/f

Making s' the subject, we have;

s' = sf/(s - f)

Since d = s + s'

Thus;

d = s + (sf/(s - f))

Expanding this, we have;

d = s²/(s - f)

The derivative of this with respect to d gives;

d(d(s))/ds = (2s/(s - f)) - s²/(s - f)²

Equating to zero, we have;

(2s/(s - f)) - s²/(s - f)² = 0

(2s/(s - f)) = s²/(s - f)²

Thus;

2s = s²/(s - f)

s² = 2s(s - f)

s² = 2s² - 2sf

2s² - s² = 2sf

s² = 2sf

s = 2f

8 0

3 years ago

Which of the following is an example of Class 3 lever system

insens350 [35]

If the fulcrum is closer to the effort, then the load will move a greater distance. A pair of tweezers, swinging a baseball bat or using your arm to lift something are examples of third class levers.

6 0

2 years ago

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Two blocks, A and B, are connected by a rope. A second rope is connected to block B and a steady, horizontal tension force of T

Oksanka [162]

Answer:

40 N

Explanation:

We are given that

Speed of system is constant

Therefore, acceleration=a=0

Tension applied on block B=T=50 N

Friction force=f=10 N

We have to find the friction force acting on block A.

Let T' be the tension in string connecting block A and block B and friction force on block A be f'.

For Block B

$T-f-T'=m_Ba$

Where $m_B$ =Mass of block B

Substitute the values

$50-10-T'=m_B\times 0=0$

$T'==40 N$

For block A

$T'-f'=m_Aa$

Where $m_A=$ Mass of block A

Substitute the values

$40-f'=m_A\times 0=0$

$f'=40 N$

Hence, the friction force acting on block A=40 N

3 0

2 years ago

A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

So

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago

Why would a soccer ball keep going forever without physics? Pls help and explain why

Pani-rosa [81]

A soccer ball would keep moving forever without physics, because without force to act upon the soccer ball, it could, or will not be able to stop the acceleration. And force is a factor in physics.

7 0

3 years ago

Read 2 more answers