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3 years ago

How does the thickness of the lens affects its focal length plz ans this i will mark as brainliest

Physics

1 answer:

Oksanka [162]3 years ago

5 0

Hello! :)

The focal length of the lens tells you how far away from the lens a focused image is created, if light rays approaching the lens are parallel. A lens with more “bending power” has a shorter focal length, because it alters the path of the light rays more effectively than a weaker lens. Most of the time, you can treat a lens as being thin and ignore any effects from the thickness, because the thickness of the lens is much less than the focal length. But for thicker lenses, how thick they are does make a difference, and in general, results in a shorter focal length.

Hope I helped and didn’t answer too late!

Good luck and stay COOL!

~ Destiny ^_^

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A heat engine operating between energy reservoirs at 20?c and 600?c has 30% of the maximum possible efficiency.

Anarel [89]

<span>The maximum possible efficiency, i.e the efficiency of a Carnot engine , is give by the ratio of the absolute temperatures of hot and cold reservoir.
η_max = 1 - (T_c/T_h)

For this engine:
η_max = 1 - [ (20 +273)K/(600 + 273)K ] = 0.66 = 66%

The actual efficiency of the engine is 30%, i.e.
η = 0.3 ∙ 0.664 = 0.20 = 20 %

On the other hand thermal efficiency is defined as the ratio of work done to the amount of heat absorbed from hot reservoir:
η = W/Q_h

So the heat required from hot reservoir is:
Q_h = W/η = 1000J / 0.20 = 5000J</span>

8 0

3 years ago

Interactive Solution 28.5 illustrates one way to model this problem. A 7.11-kg object oscillates back and forth at the end of a

____ [38]

Explanation:

Given that,

Mass of the object, m = 7.11 kg

Spring constant of the spring, k = 61.6 N/m

Speed of the observer, $v=2.79\times 10^8\ m/s$

We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :

$t_o=2\pi \sqrt{\dfrac{m}{k}} \\\\t_o=2\pi \sqrt{\dfrac{7.11}{61.6}} \\\\t_o=2.13\ s$

Time period of oscillation measured by the observer is :

$t=\dfrac{t_o}{\sqrt{1-\dfrac{v^2}{c^2}} }\\\\t=\dfrac{2.13}{\sqrt{1-\dfrac{(2.79\times 10^8)^2}{(3\times 10^8)^2}} }\\\\t=5.79\ s$

So, the time period of oscillation measured by the observer is 5.79 seconds.

3 0

3 years ago

How long (in seconds) does it take a car accelerating at 3.1 m/s2 to go from rest<br> to 51 m/s?

Nesterboy [21]

Answer:

t = 16.5s

Explanation:

Given parameters:

Acceleration = 3.1m/s²

Initial velocity = 0m/s

Final velocity = 51m/s

Unknown:

Time taken = ?

Solution:

To solve this problem we need to reiterate that acceleration is the rate of change of velocity with time.

So;

Acceleration = $\frac{v - u }{t}$

v is the final velocity

u is the initial velocity

t is the time taken

So;

3.1 = $\frac{51 - 0}{t}$

3.1t = 51

t = 16.5s

8 0

3 years ago

A celebrated Mark Twain story has motivated contestants in the Calaveras County Jumping Frog Jubilee, where frog jumps as long a

IrinaK [193]

The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

To find the answer, we need to know about the time of flight and range of projectile motion.

<h3>What's the expression of range of a projectile motion?</h3>

Range = U²× sin(2θ)/g
U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
U=√{Range×g/sin(2θ)}
Here, range= 2.20m, = 36.5°
U= √{2.20×9.8/sin(73)}

U= √{2.20×9.8/sin(73)} = 22.5m/s

<h3>What's the expression of time of flight in projectile motion?</h3>

Time of flight= (2×U×sinθ)/g
So, T= (2×22.5×sin36.5°)/9.8

= 2.73 s

Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.

Learn more about the range and time period of projectile motion here:

brainly.com/question/24136952

#SPJ1

4 0

2 years ago

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon

elixir [45]

Answer:

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity = 670 (0.7071)

Vertical component of the cannonball’s velocity = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s

6 0

3 years ago